The path of a cricket ball looks parabolic, but what about a boomerang? There are modelling assumptions which must be satisfied for the motion to be parabolic. These are:

• a projectile is a particle
• it is not powered
• the air has no effect on its motion.

Equations for projectile motion

A projectile moves in two dimensions under the action of only one force, the force of gravity, which is constant and acts vertically downwards. This means that the acceleration of the projectile is g m s−2 vertically downwards and there is no horizontal acceleration. You can treat the horizontal and vertical motions separately using the equations for constant acceleration.

To illustrate the ideas involved, think of a ball being projected with a speed of 20 m s−2 at 60° to the ground as illustrated in Figure 10.1. This could be a first model for a football, a chip shot from the rough at golf or a lofted shot at cricket.

Figure 10.1

Using axes as shown, the components are:

Using v = u + at in the two directions gives the components of velocity.

Using in the two directions gives the components of position.

You can summarise these results in a table:



The four Equations 1 , 2 , 3 and 4 for velocity and position can be used to find several things about the motion of the ball.

The maximum height:

Figure 10.2

When the ball is at its maximum height, H m, the vertical component of its velocity is zero. It still has a horizontal component of 10 m s−1 which is constant.

Equation 2 gives the vertical component as:

To find the maximum height, you now need to find y at this time. Substituting for t in Equation 4 ,

The maximum height is 15.0 m.

The time of flight:

The flight ends when the ball returns to the ground, that is when y = 0. Substituting y = 0 in Equation 4 ,

Clearly t = 0 is the time when the ball is thrown, so t = 3.46 is the time when it lands and the flight time is 3.46 s.

The range:

Figure 10.3

The range, R m, of the ball is the horizontal distance it travels before landing.

R is the value of x when y = 0.

R can be found by substituting t = 3.46 in equation 3 : x = 10t. The range is 10 × 3.46 = 34.6 m.

(Ref: Cambridge International AS and A Level Mathematics by Sophie Goldie, Series Editor: Roger Porkess, Hodder Education)