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PROJECTILE MOTION PROBLEM SOLVINGProjectile Motion Calculations

 You must be able to calculate the following quantities: horizontal and vertical components of velocity time of flight range maximum height velocity at any point

 These can all be found using the equations of motion. When using these equations, the substituted quantities must either be all horizontal or all vertical values.v=v0+at

v2=v02+2as

s=v0 t+12a t 2

vH=st

t = time (s)v0 = initial velocity (m.s-1)v = velocity at time t (m.s-1)a = acceleration (m.s-2)s = displacement (m)General Approach for solving projectile motion questions We analyse projectile motion by considering it as the vector sum of two individual, mutually perpendicular, independent motions- one in the horizontal direction and the other in the vertical.Because these are both linear motions and because the acceleration is constant in each direction (zero in the horizontal direction & g= 9.8 m/s2 down in the vertical. We can use Newton’s equation of motion to perform this analysis.Basically

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To analyze the motion of a projectile, we consider its horizontal and vertical motion separately. Horizontal Vertical Acceleration is 0 m.s-2

Horizontal component of velocity vH is constant

Acceleration is always 9.8 m.s-2

Vertical component of velocity vv at the highest point is 0 m.s-1

Horizontal Calculations: As we know, displacement is equal to velocity multiplied by time.So in the context of projectile,VH = S H ÷ tVertical Calculations:The two relevant equations here are:1)v=v0+at

2¿ s=v0t+12a t2

3¿2as=v2−v02

Within the context of Projectlie motion problems The first relationship is used to calculate the vertical component of the velocity at any time and the time of flight of the projectlie The second equation may be used to calculate the time of flight of the projectlies

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Thrid equation is the most straightforward way of calculating the maximum height achieved by the projectile during the flight.

Standard Projectile Motion Calculations In analysing a projectile motion we need to perform the following calculations:

Determine the time of flight of the projectileIn the case of horizontal projection : we find the time of flight by just finding the time taken for a free falling body to fall from the launche height vertically down to the ground.In the case of angled projection- we find the time taken for the projectile to attain its maximum height. By the symmetry of this motion, the time of flight is just double this time.

Determine the velocity of the body (body magnitude and direction) at any point in its flight.We find the vertical and horizontal components of the velocity at any time, and then add these components together vectoially. This can be done either by drawing a scale diagram, or by using Pythagoras theorem and right angled triangle trignometry.

Determine the range of the projectile.This is found by multiplying the horizontal velocity by the time of flight.

Determine the maximum height attained by the projectile.At its maximum height the vertical component of the velocity is zero. We can calculate the maximum height (launch height) attained by the projectile by substituting Vy = 0 into the equation.

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SAMPLE CALCULATIONSThere are five different projectile motion scenarios for which we can perform the calculations listed above. They are :

1. An object is launched horizontally from some height and then falls to ground in parabolic path.

2. An object is projected at some angle above the horizontal, and finishes at the same height.

3. An object is projected at some angle above the horizontal, but the object finishes at a levelBelow the initial launch height.

4. An object is projected at some angle above the horizontal , but the object finishes at a level above the initial launch height.

5. An object is projected from a height at some angle below the horizontal , ie it is thrown downwards

Horizontal projectionExamples of this type of motion are:

a ball rolling off a table top. Supplies being dropped from an aeroplane to a life raft at

sea. A stone thrown from a cliff top A car plunging off a cliffMethod Scenario 1 :

Because all the vertical motion is in the downward direction , call this direction positive.

Establish the initial conditions a= 9.8 Vy = 0, Vx=V where V is the initial speed

Time of flight t is a vertical calculation using s=v0 t+ 12 a t2

with Vy = 0 Range is a horizontal calculations using S=Vh times t To find the velocity v at any time t, find the vertical

velocity at that time, using v=v0+atThen , add the vertical velocity Vy and horizontal velocity Vx vertically, to find the velocity v.

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EXAMPLE 1

A coin is dropped into a wishing well. Find the vertical velocity after 3 s.

Use: v⃗=v⃗0+ a⃗ t

EXAMPLE 2

How deep is the well if the coin hits the bottom at t=4 s?

Use: s⃗= v⃗0 t+12a⃗ t 2

EXAMPLE 3

An arrow fired horizontally from a cliff 122.5 m high. Its initial velocity is 10 ms-1 horizontally. After 3.0 s, the vertical velocity is 29.4 ms-1. Assume no air resistance. Find the velocity ( v⃗¿at 3.0 s.

v⃗H=10ms−1

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h = 122.5 m

TIME OF FLIGHTThe time of flight is the time taken for the object to fall vertically to the ground.

A B C

Time of flight is influenced by:

launch height launch angle strength of ‘g’ at a particular location air resistance

A, B, C all fall the same distance A, B, C have the same time of flight A, B, C hit the ground at the same time.

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initial velocity

EXAMPLE 4 HORIZONTAL LAUNCH

A rescue plane at 120 m drops a food parcel. How long before the parcel hits the ground?

EXAMPLE 5 GROUND LEVEL LAUNCH

A cricketer in the outfield throws the ball with a velocity of 25 ms-1 at an angle of 30° above the horizontal.Find the time of flight.

RANGE

Range is the horizontal distance from the point of launch to the point of landing.

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Using: displacement=velocity×time

range= v⃗h×time time=time of flightv⃗h=v⃗ cosθ

To increase the range:

increase v⃗h

or

increase time of flight

EXAMPLE 6

Jim hits a golf ball with a velocity of 30 ms-1 at an angle of 15° to the horizontal. Find:(i) time of flight

(ii) range

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PRACTICE QUESTIONS A mortar shell is fired from ground level with a velocity of 100 m.s-1 at 80° above the horizontal. Find:

a) the initial horizontal and vertical components of velocity

b) the vertical component of velocity 13 s after firing

c) the magnitude and direction of the velocity of the mortar shell after 13 s

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d) the magnitude and direction of the velocity of the shell at the top of its flight

e) the maximum height reached by the shell

f) the shell's time of flight

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g) the range of the shell

ExampleA parcel is dropped from a plane into the sea. The plane isflying at 100 m.s-1 horizontally at 120 m above sea level.Find:

a) the vertical distance fallen by the parcel after 2 s

b) the vertical component of velocity of the parcel after 2 s

100m. s−1

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c) the magnitude and direction of the velocity of the parcel after 2 s

d) how long the parcel takes to hit the water

ExampleYou throw a baseball 42.8 m and it spends 1.64 s in the air. Calculate the magnitude and direction of the ball’s final velocity just before it is caught (assuming it is caught at

42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m42.8 m

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the same height as it was thrown). Hence, state its initial velocity.

TRY out these questions 1) An aeroplane is to drop a parcel of medical supplies to a

boat in the sea. The aeroplane is flying horizontally at a height of 240 m above sea level with a speed of 90ms-1.Calculate : a) the time between the parcel leaving the aeroplane and hitting the water. ( ie the time of flight)b) How far from the boat the aeroplane must release

the parcel ( ie the range of the projectile.c) The velocity of the parcel on hitting the water.

2) A cricketer, fielding in the outfield, throws a ball to the wicket keeper, with a speed of 23ms-1 at an angle of 30 deg above the horizontal. The wicket keeper catches the ball at the same level at which it was originally thrown. Find : a) The time of flight of the ballb) The distance of the fielder from the wicket keeper

(range)c) The velocity with which the ball strikes the wicket

keeper’s gloves.

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The Effect of Launch Angle on Range

To increase the range, you can increase the horizontal component of velocity vH or the time of flight t.

to increase vH, the launch angle must be smaller

to increase t, the vertical component of velocity must be increased and hence the launch angle must be larger

Maximum range is achieved when these two factors are balanced – the projectile is launched at 45 °.

If two launch angles add up to 90 °, they will produce the same range.

45 °θθ

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The Effect of Air Resistance

Air resistance acts in the opposite direction to the object’s motion. This force depends on the object’s:

size -

shape and surface texture -

speed -

and on the density of the air (denser air means more air particles will contact the object’s surface).

Without air resistance

(parabolic path)

Horizontal displacement (m)

Height above ground (m)

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Air resistance affects both the horizontal and vertical components of velocity.

The horizontal component of velocity decreases over the projectile’s flight. Since the range is directly proportional to vH, the range decreases.

The vertical component of velocity decreases more rapidly as the projectile travels upwards since both air resistance and gravity are acting against it. Therefore, it reaches zero velocity sooner and the time to reach maximum height is decreased. When the projectile is falling back down, air resistance opposes its motion and increases the time taken for it to fall from maximum height. Overall, the total time of flight is slightly reduced.

OR MORE ON AIR RESISTANCE Air resistance will oppose the velocity of the projectile.

MOVING UP Air resistance assists gravity (- in effect, increasing downward acceleration).

the projectile reaches maximum height sooner (- time to reach max. height is less) and

does not go up as far.

FALLING Air resistance opposes gravity (- in effect, decreasing downward acceleration).

the projectile takes slightly longer to fall.

OVERALL the net effect of air resistance is to slightly reduce time

of flight (almost the same). If anything, it will decrease the range.

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APPLICATION: Projectiles in Sport

To achieve maximum range, we need to consider:

effect of launch angle effect of the launch height the role of air resistance

You would expect the maximum range for a uni-level launch to be with a launch angle of 45°.

(Check value for horizontal velocity for different angles using v⃗H= v⃗ cosθ¿

1. EFFECT OF LAUNCH ANGLE:

2. EFFECT OF LAUNCH HEIGHT:

launch height

For a given initial velocity:

The greater launch height means the projectile is in the air for longer (has further to fall)

v⃗H acts for longer causing a greater range

ground level

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A projectile launched from above ground level, with an angle of slightly less than 45°, it will travel a shorter horizontal distance (in returning to its launch height) but it will travel further from this height to the ground for the same initial speed. See diagram below.

Horizontal velocity (with smaller launch angle) is greater than if launched at 45° ( v⃗H= v⃗ cosθ¿

Vertical velocity (due to smaller launch angle) on returning to launch height is smaller (takes longer to fall remaining distance)

Both these factors increase range. The object is in the air for longer while moving horizontally.