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Linear Motion with constant acceleration Physics 6A Prepared by Vince Zaccone For Campus Learning Assistance Services at

Linear Motion with constant acceleration Physics 6A Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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Linear Motionwith constant acceleration

Physics 6A

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

As long as its acceleration is constant*, the motion of any

object can be described by the following simple formulas:

When using these formulas it is important to be clear about your variables.Especially important is where and when the position and time are zero.

A couple of examples will help clarify this.

*both the magnitude and direction must be constant

020

2

0

221

00

xxa2vv

atvv

attvxx

(1)

(2)

(3)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

By definition, average acceleration is just (change in velocity)/(change in time).Since the car sped up from 0m/s to 30m/s in 5.6 seconds, this is a simple calculation:

2sm

s

sm

4.56.5

30a

This is the answer for part a)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Another way to do part a) is to use formula (2) from before:

atvv 0

In this case we are starting from rest, so v0 = 0m/s,and our final speed is v = 30m/s.

The elapsed time is t = 5.6s

Putting these numbers into the equation we get:30m/s = 0 + a(5.6s)

Solving for a gives us the answer:a ≈ 5.4m/s2

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Now that we have the acceleration from part a), we can get the answer to b)Which of our 3 equations is appropriate here?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

020

2

0

221

00

xxa2vv

atvv

attvxx

(1)

(2)

(3)

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Now that we have the acceleration from part a), we can get the answer to b)Which of our 3 equations is appropriate here?

Equation (1) or (3) will work in this case.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

020

2

0

221

00

xxa2vv

atvv

attvxx

(1)

(2)

(3)

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Using Equation (1):

m

2ss

m21

221

00

85x

6.54.500x

attvxx

2

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Using Equation (3):

m

sm

sm

sm2

sm

020

2

83x

x8.10900

0x4.52030

xxa2vv

22

2

2

Note: The discrepancy is due to round-off error from the original calculation of a

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Now we move on to the braking part of the problem:We can organize our information in a table if we want:

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

x0 = x =

v0 = v=

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

x0 =0 x =50m

v0 =30m/s v= 0

Now we move on to the braking part of the problem:We can organize our information in a table if we want:

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

020

2

0

221

00

xxa2vv

atvv

attvxx

(1)

(2)

(3)

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Compare this information with our formulas from earlier.Which formula is the easiest one to use to find acceleration?

Now we move on to the braking part of the problem:We can organize our information in a table if we want:

x0 =0 x =50m

v0 =30m/s v= 0

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

020

2

0

221

00

xxa2vv

atvv

attvxx

(1)

(2)

(3)

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Now we move on to the braking part of the problem:We can organize our information in a table if we want:

x0 =0 x =50m

v0 =30m/s v= 0

Compare this information with our formulas from earlier.Formula (3) gives us what we need.

Here’s the calculation:

2sm

m2

sm

020

2

9a

050a2300

xxa2vv

Why did this come out negative?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Now that we have acceleration, we can find the time for part d)

Which formula should we use?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

020

2

0

221

00

xxa2vv

atvv

attvxx

(1)

(2)

(3)

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Now that we have acceleration, we can find the time for part d)

Which formula should we use?

Either of (1) or (2) will work, so choose the easier one: Formula (2)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

020

2

0

221

00

xxa2vv

atvv

attvxx

(1)

(2)

(3)

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Here is the calculation:

ss

sm

sm

0

3.39

30t

t)9(300

atvv

2

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #1A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.

Find the following:

a) the average acceleration while it is speeding up

b) the distance traveled while speeding up

c) the average acceleration while braking

d) the time elapsed while braking

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

y=0

y=2.1m

First we need to set up a coordinate system. A convenient way to do it is to let the lowest point be

y=0, then call the upward direction positive.

With this choice, our initial values are:

Y0 = 2.1m

V0 = 30m/sV0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

For vertical motion on the surface of Earth, the acceleration will always be downward, with

magnitude g = 9.8m/s2

Important: when you see g in a formula, it will be a positive number. Every time. No exceptions.

V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

Now we can think about the problem:

What is special about the maximum height?

In other words, do we know anything specific about the ball when it reaches the maximum height?

V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

Now we can think about the problem:

What is special about the maximum height?

In other words, do we know anything specific about the ball when it reaches the maximum height?

YES! This is the most important piece of information in this type of problem:

the vertical velocity is zero at the top.

V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

Now we can use a formula to find the max height.

Here they are again, adjusted for vertical motion:ymax=?

v = 0 here

020

2

0

221

00

yyg2vv

gtvv

gttvyy

(1)

(2)

(3)

V0 = 30 m/s Which one matches our information?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

Now we can use a formula to find the max height.

Here they are again, adjusted for vertical motion:ymax=?

v = 0 here

Which one matches our information?

We don’t have information about the time, so we use Equation (3)

020

2

0

221

00

yyg2vv

gtvv

gttvyy

(1)

(2)

(3)

V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

Here’s the calculation:ymax=?

v = 0 here

mmax

mmaxsm2

sm

020

2

48y

1.2y8.92300

yyg2vv

2

V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

We can use (2) to find the time to get to the top(we could have done this before part a) if we had wanted)

ymax=48mv = 0 here

s

sm

sm

0

1.3t

t8.9300

gtvv

2

V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

Where is the ball when it is traveling at its maximum speed?

v = 0 here

V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

ymax=48m

Where is the ball when it is traveling at its maximum speed?

Just before it hits the ground. A good way to think about this is to realize that it is speeding up as it falls, so the farther it falls, the faster it is moving.*

Note: as the ball passes by the starting point it has a velocity of 30m/s downward. This is just the

opposite of the initial velocity! In fact, at any height, the ball will be moving at the same speed on the

way up or the way down.

*We are ignoring air resistance in this problem, so there is no terminal speed.y=0

y=2.1m

ymax=48mv = 0 here

V0 = 30 m/s

Fastest here

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

We can use formula (3) to find the speed just as the ball hits the ground. Our final position will be

y = 0

We can use anywhere for the starting point, as long as we put in the appropriate initial velocity.

y=0

y=2.1m

ymax=48mv = 0 here

V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

Here is the answer in 2 ways:

With y0=2.1m:ymax=48m

v = 0 here

sm

msm2

sm2

020

2

7.30v

1.208.9230v

yyg2vv

2

sm

msm2

020

2

7.30v

4808.920v

yyg2vv

2

With y0=48m:V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

Time for the final question. Which formula should we use to get the time when the ball hits?

y=0

y=2.1m

V0 = 30 m/s

ymax=48mv = 0 here

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

Time for the final question. Which formula should we use to get the time when the ball hits?

We can actually use either (1) or (2), but we have to be careful!

Let’s try it first with Equation (1)…

y=0

y=2.1m

V0 = 30 m/s

ymax=48mv = 0 here

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

We need to put in y,y0,v0 and g

We know it lands on the ground, so y = 0

If we use y0 = 2.1m and v0 = 30m/s:

2sm

21

smm

221

00

t8.9t301.20

gttvyy

2

How do we solve this???

y=0

y=2.1m

V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

We need to put in y,y0,v0 and g

We know it lands on the ground, so y = 0

If we use y0 = 2.1m and v0 = 30m/s:

2sm

21

smm

221

00

t8.9t301.20

gttvyy

2

If we want to solve this one, we need to use the quadratic equation. This will work just fine, but it is more work than we should really need to do here.

We get 2 answers:

t = 6.2s and t = -0.07sy=0

y=2.1m

V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

We need to put in y,y0,v0 and g

We know it lands on the ground, so y = 0

If we use y0 = 2.1m and v0 = 30m/s:

2sm

21

smm

221

00

t8.9t301.20

gttvyy

2

If we want to solve this one, we need to use the quadratic equation. This will work just fine, but it is more work than we should really need to do here.

We get 2 answers:

t = 6.2s and t = -0.07s

Choose the positive answer in this case.

y=0

y=2.1m

V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

That took some tedious algebra to find the answer. How can we avoid this?

v = 0 here

y=0

y=2.1m

V0 = 30 m/s

ymax=48m

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

That took some tedious algebra to find the answer. How can we avoid this?

We can try using a different starting point.

How about starting at the top?

Putting in y0 = 48m and v0 = 0:

v = 0 here

2sm

21m

221

00

t8.90480

gttvyy

2

This time it is much easier to solve!y=0

y=2.1m

V0 = 30 m/s

ymax=48m

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

We get 2 answers again:

t = 3.1s and t = -3.1s

Again we choose the positive value, but does it actually answer our question?

v = 0 here

y=0

y=2.1m

V0 = 30 m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

We get 2 answers again:

t = 3.1s and t = -3.1s

Again we choose the positive value, but does it actually answer our question?

NO! This is only the time it takes to fall from the maximum height. We need to add it to the time it

took to get to that max height. We found that earlier – it was 3.1s

So our answer is the total time of 6.2s, as before.

ymax=?v = 0 here

y=0

y=2.1m

V0 = 30 m/s

ymax=?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

Now we can do it one more time, but using Equation (2) instead.

We already found the final speed to be v = 48m/s

Here is the calculation:

ymax=?v = 0 here

s

sm

sm

sm

0

2.6t

t8.9307.30

gtvv

2

y=0

y=2.1m

V0 = 30 m/s

ymax=?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

y=0

y=2.1m

V0 = 30 m/s

Now we can do it one more time, but using Equation (2) instead.

We already found the final speed to be v = 48m/s

Here is the calculation:

ymax=?v = 0 here

s

sm

sm

sm

0

2.6t

t8.9307.30

gtvv

2

IMPORTANT NOTE: make sure you put in the correct final velocity here. It is negative because

the ball is moving downward!

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #2A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.

Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:

a) What is the maximum height reached by the ball?

b) When does it achieve that maximum height?

c) What is the maximum speed achieved by the ball?

d) When does the ball hit the ground?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #3

A model rocket is fired vertically from a launch pad on the ground. The engine provides an upward acceleration of 5 m/s2 for a time of 10 seconds. After that time the rocket is in free fall. Assume that the parachute fails to deploy, and the rocket eventually plummets to the ground.

Sketch graphs of the rocket’s vertical height vs. time and its velocity vs. time.

Find the maximum height of the rocket.

When does it land, and how fast is it moving just before it hits the ground?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Sample Problem #3

A model rocket is fired vertically from a launch pad on the ground. The engine provides an upward acceleration of 5 m/s2 for a time of 10 seconds. After that time the rocket is in free fall. Assume that the parachute fails to deploy, and the rocket eventually plummets to the ground.

Sketch graphs of the rocket’s vertical height vs. time and its velocity vs. time.

Find the maximum height of the rocket.

When does it land, and how fast is it moving just before it hits the ground?

Answers:

Max Height = 378 m

Time to impact = 23.9 s

Impact speed = 86 m/s