Projectiles
Introduction

Discus, javelin, shot, and hammer throwing are Olympic field events involving projectiles. Other Olympic events involve jumping. For example, while competitors in the long jump frequently hurl themselves over 7 m into a sandpit, ski jumpers can travel over 100 m between take-off and landing.

The distance that a competitor can throw a projectile depends upon the strength and technique of the athlete as well as the aerodynamics of the object projected. The speed of the object before release, the angle at which it is released, and the ability to maximize the aerodynamic features of the projectile all contribute to the distance between the release and landing positions. Rotation of a javelin or discus in flight produces extra lift which prolongs flight and therefore results in longer distances.

In this consideration of projectile motion we will treat all thrown objects as small point masses and will ignore the effects of the air through which the projectile is travelling. The horizontal distance travelled by a fired or thrown object is called the range and we will show how this can be determined using the equations of motion.

Modelling projectile motion
The apparatus shown in Fig.1 can be used to drop the blue ball vertically while simultaneously projecting the orange ball horizontally.

 Figure 1. Simultaneous release.
The projected ball and the ball dropped vertically hit the floor at …

The result of Fig.1 seems to defy common sense. It would seem to suggest that an archer, firing a bolt horizontally from a crossbow while at the same time dropping the crossbow, would see both the bolt and the crossbow landing on the ground at the same instant!

In practice, they would not hit the ground at the same instant because it would be difficult to synchronize releasing the bolt with dropping the bow. Additionally, the flights on the bolt are designed to give it 'lift', keeping it in the air longer. Many other projected objects adopt similar techniques to prolong flight.

 Figure 2. Designed to give lift.

In our simple model for the motion of a projectile, we have to make the following assumptions:

• Only the force of gravity acts on the vertical motion.

• No frictional forces impede the horizontal motion.

The blue ball is dropped vertically. In equal time intervals the distance it travels increases, so the ball is accelerating as it falls. The blue ball is freely falling due to the force of gravity, so its
acceleration
An object's acceleration is its rate of change of velocity.
acceleration
is a constant 9.81 ms−2.

Approximately how far does the blue ball fall in the first 0.1 seconds after it is released?
Approximately how far does the blue ball fall in the first 0.2 seconds after it is released?

The orange ball is projected horizontally and during its flight it travels both downwards and to the left. Its motion has both horizontal and vertical components.

Considering only the downward component of motion, we can see from the positions shown in Fig.1 that the vertical displacements of the projected ball, at any instant, are identical to those of the dropped ball. Therefore the projected ball is also accelerating downwards at 9.81 ms−2.

The horizontal distance between successive positions is equal, so the horizontal motion is a constant speed.

The ball is projected horizontally with a speed of 5 ms−1. What horizontal distance does the ball travel in the first 0.1 seconds?
The ball is project horizontally with a speed of 5 ms−1. What horizontal distance does the ball travel in each subsequent 0.1 second interval?

The movement of the projected ball can be treated as being made up from independent horizontal and vertical motions.

• The horizontal motion is a constant speed.

• The vertical motion is a constant acceleration.

Horizontal projection
The motorcycle in Fig.3 is travelling horizontally at take-off. In flight it follows a projectile path, so we can treat its vertical and horizontal motions separately. To reach the landing-zone the bike must travel 48 m horizontally in the same time as it travels 19.62 m vertically. Use the animation to determine the horizontal speed required to land in the centre of the landing-zone.

The horizontal speed required to land in the centre of the landing-zone is …
 Figure 3. Safe landing!

 The horizontal speed of the motorcycle is constant throughout its flight, so the time taken for the flight can be determined because …
 the horizontal speed required to land in the centre of the landing-zone is 24 ms−1.

Looking at its vertical motion, the motorcycle has been falling with an acceleration of 9.81 ms−2 for a time of 2 seconds when it lands in the centre of the landing-zone.

 We can determine the distance travelled vertically using the second equation of motion.
 Initially the bike has no vertical component to its velocity. Therefore in the 2 seconds of its flight, the vertical distance travelled is

Since the calculated horizontal and vertical distances travelled match those from the situation in Fig.3, we can see that treating the projectile's motion as separate horizontal and vertical motions accurately describes the situation.

At this point it is interesting to note that the flight time depends only on the vertical distance between take-off and landing and is independent of the take-off speed.

During the fall lasting 2 seconds, the motorcyclist travels horizontally at a constant speed covering a horizontal distance of 48 m, so the horizontal speed is 24 ms−1. Complete the following statements.

• Take-off speeds above 24 ms−1 mean that the motorcyclist covers 48 m horizontally in the 2 seconds of the flight and so will land the landing zone.

Take off speeds below 24 ms−1 mean that the motorcyclist covers 48 m horizontally during the flight and so the landing point will be the landing zone!

In this situation we have modelled the motion of the projectile as separate horizontal and vertical motions and have used the equations of motion to determine the horizontal and vertical displacements at one particular time.
Projection at an angle
Set the
velocity
An object's velocity states both the speed and direction of motion relative to a fixed reference point.
velocity
of the cannon ball in Fig.4 to 13.5 ms−1 and note the effect of altering the projected angle on the range.

 Figure 4. Altering the angle.
Complete the following statements to summarize the behaviour of the ball in Fig.4.

• For any set angle of projection, increasing the firing speed the distance travelled.

For any set speed, the range initially as the angle increases but if the angle gets too large.

When the cannon is fired, the ball leaves with a specific velocity. Being a
vector
A vector quantity is specified fully only when its magnitude and direction are both quoted.
vector
quantity, the velocity has both size and direction.

 Figure 5. The projected speed is v and the angle θ.

The velocity has both horizontal and vertical components. The horizontal component of the velocity vH is given by

The vertical component of the velocity vV is given by

Our simple model of projectile motion states that the speed in the horizontal plane is constant, so for the entire duration of the flight the horizontal speed of the flight is where v is the projected speed and θ the angle.

Set the velocity of the cannon ball in Fig.4 to 13.5 ms−1 and the angle of projection to 45°. Note the arrow in the diagram showing the components of the speed throughout the motion. While the horizontal speed of the ball in Fig.6 is constant, the speed in the vertical direction changes.

 Figure 6. Cannonball with velocity arrows.
Complete the following statements to summarize how the vertical component of the velocity changes during the flight.

• The speed in the vertical direction until the projectile reaches its . At the top of its motion its vertical speed is 0 ms−1. As it starts to fall again, the downwards speed .

Initially, the vertical speed of the cannon ball decreases to zero and then increases again during the downward journey. As gravity is the only force acting in the vertical direction, its vertical motion is a
deceleration
The deceleration of an objects is a measurement of its rate of change of velocity as it slows down.
deceleration
from its initial value to 0 ms−1 at approximately 10 ms−2 as it rises to its highest point. On the downward part of its journey, the vertical speed increases from 0 ms−1 at 10 ms−2.

In the following example, a ball is projected with a velocity of 25 ms−1 at an angle of 30° to the horizontal.

 The horizontal component of the velocity .
 vH = 25 cos 30;
 vH = 21.65 ms−1
 The initial vertical component of the velocity
 vV = 25 sin 30
 Or, vV = 12.5 ms−1
 The time taken to reach maximum height can be calculated, assuming that g = 10 ms−2.
 At the highest point, the vertical speed is 0 ms−1 so we can use the first equation of motion to calculate the time taken to decelerate from 12.5 ms−1 to 0 ms−1.
Information required:
 u v a s t

Collecting values:
 u v a s t 12.5 0 −10 not required ?

 Substituting into the equation of motion:
 From Fig.6 it is clear that the motion of the ball is symmetrical and so the time to fall from its maximum height back to the ground will be the same as the time to reach the maximum height. In this example the total flight time will therefore be 2.5 seconds.
 During the whole 2.5 seconds of the flight the ball had a constant horizontal speed of 21.65 ms−1. Therefore the range can be calculated from Range = speed × time
 Range = 21.65 × 2.5 Range = 54.1 m

By regarding the horizontal and vertical motion as separate and using the equations of motion we can determine the range of a fired projectile.

How far will it go?
In Fig.7 the projectile's initial velocity is fixed. Alter the angle of projection to find the angle required for maximum range.

 Figure 7. Finding the best angle.
The angle required for maximum range in Fig.7 is …

Fig.7 illustrates clearly that the range of the projectile depends on the angle of projection. Maximum range is achieved with a projection angle of 45°. This observation can be proven mathematically as follows.

 For the general case of an object projected with a speed of v at angle θ to the horizontal in an area where the acceleration due to gravity is g:
 Range = Horizontal speed × flight time
 The flight time = 2 × time to reach highest point
 Therefore, Range = Horizontal speed × 2 × time to reach highest point
 The horizontal speed is vcosθ throughout the motion.
 Therefore, range = 2t × vcosθ
We can find the time to reach the highest point, t, by considering the vertical motion only. Vertically,
 v u a s t 0 vsinθ −g not required t

Since v and g are constants for this situation, the maximum range will be achieved when the sin 2θ is maximum. The maximum value of a sine function is 1 and this happens for angles of 90°. If 2θ = 90°, then θ = 45°.

This mathematical derivation confirms the observations from Fig.7. We must remember, however, that our model for the motion of the projectile depends upon certain assumptions and is valid only while these assumptions hold.

Summary

The range of a projectile depends upon the speed before release and the angle of projection.

Projectile motion can be modelled by considering the speeds of the separate horizontal and vertical components of the motion.

The speed in the horizontal direction is constant.

The projectile's vertical motion is a uniform acceleration caused by the force of gravity.

The maximum range for any given speed is achieved for a projection angle of 45°.

The actual velocity at any instant during a projectile's flight is the vector sum of the horizontal and vertical speeds.

Exercises
1. Fig.8 shows a piece of apparatus used to illustrate some of the principles of projectile motion. A steel ball is held by an electromagnet 78.4 cm above a catching mechanism.

• Calculate the time the ball takes to fall from the electromagnet into the catcher.
s   (to 1 d.p.)

In another experiment, the electromagnet and the catcher are mounted on top of a trolley which is placed on a horizontal frictionless track. The trolley is pushed and released so that it moves horizontally with a speed of 2 ms−1. When the trolley passes point A, the electromagnet releases the ball.

How long does the ball take to fall 78.4 cm?
s   (to 1 d.p.)

What is the initial horizontal velocity of the ball immediately prior to its release?
ms−1   (to the nearest whole number)

The ball lands in the catcher. How far is the catcher from point A when the ball first lands in it?
m   (to 1 d.p.)
 Figure 8.
2. The teacher demonstrating the apparatus shown in Fig.8 states that the experiment shows that the motion of a projectile can be treated as independent horizontal and vertical motions.

Which of the following statements most accurately describes the horizontal motion of the projectile?
3. Which of the following statements most accurately describes the vertical motion of a projectile launched by the apparatus in Fig.8?
 Figure 9.
4. A teacher uses the apparatus shown in Fig.9 to demonstrate some of the principles of projectile motion. The launching mechanism, mounted on a trolley, fires a ball directly upwards with an initial velocity of 3.9 ms−1.

• Calculate the height the ball will rise above the launching mechanism when fired.
cm   (to 1 d.p.)

Calculate the time taken for the ball to rise to its highest point above the release point.
s   (to 1 d.p.)

The trolley is pushed so that it travels at a steady speed of 1.1 ms−1. When the trolley passes point A, the mechanism fires the ball vertically with an initial speed of 3.9 ms−1. Calculate the distance travelled by the trolley before it catches the ball.
m   (to 2 d.p.)

The ball is caught 2 m to the right of the launch point. Calculate the initial speed of the trolley.
ms−1   (to 1 d.p.)
5. A motorcyclist jumping across a ravine, as shown in Fig.10, lands in the middle of the landing-zone which is 23.7 m vertically below the take-off point. The centre of the landing-zone is 38.5 m horizontally away from the take-off point.

• Calculate the time between take-off and landing.
s   (to 1 d.p.)

What is the minimum speed at take-off that will ensure that the motorcyclist lands in the centre of the landing-zone?
ms−1   (to 1 d.p.)

On one attempt, the motorcyclist has a take-off speed of 30 ms−1. How far from the centre of the landing-zone does the cyclist land?
m   (to 1 d.p.)
 Figure 10.
6. In an effort to jump further, the motorcyclist suggests that he could use a heavier, more powerful motorcycle. He is concerned that this heavier machine would 'drop more quickly'. What advice would you give the motorcyclist?
 Figure 11.
7. A stunt car driver uses a ramp, as shown in Fig.11, to jump over a parked bus. The path of the car is as shown (a=9.8ms−2). The horizontal and vertical components of the car's velocity at take-off are as shown.

• What is the time taken for the car to reach its highest point after leaving the ramp?
s   (to 2 d.p.)

What is the time taken for the car to reach the landing area?
s   (to 2 d.p.)

Calculate the horizontal distance between the landing point and the end of the take-off ramp.
m   (to the nearest whole number)
 Figure 12.
8. A golfer practising her drive stands on the edge of a cliff shown in Fig.12 and hits a golf ball horizontally. The sea is 37 m vertically below the edge of the cliff. After one particular swing the ball follows the path as shown in Fig.12 before landing in the water.

• How long after it is struck does the ball hit the water?
s   (to 2 d.p.)

What is the horizontal component of velocity of the ball during its motion?
ms−1   (to the nearest whole number)

What is the vertical component of the ball just before it enters the water?
ms−1.

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