1.) An object projected straight upward with an initial velocity of 88.2m/s will reach its maximum altitude of 396.9m in 9s. If after 5s the object is at an altitude of 318.5m moving upward, it will be at an altitude of 318.5m moving downward after:

5s is 4s before reaching the maximum altitude. Therefore the object will be at the same altitude moving downward 4s after reaching the maximum altitude.

9s+4s=13s

2.) If an object is thrown straight up, when it reaches its maximum altitude its displacement will be:

smax =vitmax+12 gtmax

2

tmax =0−vig =

−vig

smax =vi−vig

⎛ ⎝ ⎜

⎞ ⎠ ⎟+

12 g

−vig

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

smax =−vi

2

g + 12 g

vi2

g2 ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

smax =−vi

2

g + 12

vi2

g

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

smax =−vi

2

2g

The maximum altitude depends on the initial

velocity.

The diagram below shows consecutive pictures taken at equal time intervals of an object moving with an increasing velocity.

3.) The direction in which the object moved was:

4.) The direction of the object's acceleration was:

To the left

To the left

5.) When a 10 kg stone is dropped into a well the splash is heard 8 seconds later. If a 5 kg stone is then dropped into the well its splash will be heard after:

Acceleration during free-fall is independent of the object’s mass and the speed of sound is constant. Therefore the time between dropping the object and hearing the splash does not depend on the object’s mass.

The splash is heard after the object reaches the bottom of the well and the sound of the splash travels back to the top of the well.

s =vit1 +12 gt1

20

t1 = 2sg

t2 = svsound

ttotal =t1 + t2

6.) If the speed of sound in air is 330m/s the depth of the well in question #5 is:

The displacement of the falling object is equal in magnitude and opposite in direction to the displacement of the sound.

s1 =−s2

12 gt1

2 =−vsoundt2

t1 + t2 =8

t2 =8−t1

12 gt1

2 =−vsound8 −t1( )

−4.9t12 = −330 8 − t1( )

4.9t12 =2640 −330t1

4.9t12 + 330t1 −2640 =0

a =4.9b=330c=−2640

t1 =−b± b2 −4ac

2a

t1 =−330 ± 330( )2 −4 4.9( ) −2640( )

2 4.9( )

t1 =−330 ±400.8

9.8

t1 =−330 +400.8

9.8 =7.22s

t1 =−330 −400.8

9.8 =−74.6s

t2 =8−t1t2 =8s−7.22s

t2 =0.78s

s =vsoundt2

s= 330ms( ) 0.78s( )

s=257m

7.) After a loud sound is made the echo from a distant cliff is heard 12s later. If the speed of sound in air is 330m/s, the distance to the cliff is:

The echo is heard after the sound has traveled from its source to the cliff and its reflection has returned. Therefore the time for the sound to travel to the cliff is 6s.

s =vsoundt

s= 330ms( ) 6s( )

s=1980m

A rocket car is propelled along a long straight track by an engine that can produce an acceleration of 50m/s2.The car starts from rest. The rocket engine is turned on for 5s and then turned off. 3s after the engine is turned off a parachute is deployed which produces an acceleration of -20m/s2

Part OneAccelerated motion

s =? v=?

vi =0

vf =?

a=50 ms2

t=5s

vf =vi + at

vf =0 + 50 ms2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 5s( )

vf =250 ms

s =vit+12 at

2

s=0 + 12 50m

s2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟ 5s( )2

s=625m

Part TwoConstant velocity

s =?

v=250mst=3s

s =vt

s= 250 ms( ) 3s( )

s=750m

Part ThreeAccelerated motion

s =? v=?

vi =250msvf =0

a=−20 ms2

t=?

t =vf −via

t=0−250ms−20m

s2

t=12.5s

v =vi −vf2

v=250ms +0

2

v=125ms

s = vt

s= 125ms( ) 12.5s( )

s=1562.5m

stotal =625m+ 750m+1562.5m

stotal=2937.5m

t total =5s+ 3s+12.5s

ttotal=20.5s

v =stotalttotal

=2937.5m20.5s

v=143.3ms