  # A projectile problem

A boy stands on the edge of a 100m high cliff and shoots a projectile (A) vertically upwards at 30 ms-1. One second later his friend at the base of the cliff shoots a second projectile (B) vertically upwards at 50 ms- 1.

Determine:
(a) the height above the ground at the base of the cliff where the two projectiles pass each other
(b) the time after the projectile A was launched that this happens
(c) the velocities of projectile A and B when they pass each other

Let the time after projectile A is launched until they pass each other be t and let the height above the ground when this happens be h.

Take g = 9.8 ms- 2.

For projectile A: h = 100 + 30t + ¢ gt2 equation 1
For projectile B: h = 50(t-1) + ¢ g(t-1)2 equation 2
Therefore: 100 + 30t + ¢ gt2 = 50(t-1) + ¢ g(t-1)2 = 50t û 50 + ¢ x 9.8 x (t2 û 2t +1)
So remembering that g will be negative if the velocities upwards are positive:
100 + 30t + ¢ gt2 = 59.8t û 54.9 + ¢ gt2
154.9 = 29.8t and so t = 5.2 s

Substituting for t in equation 1 gives: h = 100 + 30x5.2 û ¢ x9.8x5.22
h = 255.91 - 132.39 = 123.5 m

The projectiles pass each other when they are 123.5 m above the ground

Checking this using equation 2 also gives h = 123.5 m

### Velocity of A at h = 123.5 m

v = u ûgt = 30 û 9.8x5.2 = -20.1 ms-1 A is travelling downwards

### Velocity of B at h = 123.5 m

v = u ûgt = 50 û 9.8x4.2 = +8.84 ms-1 B is travelling upwards

A VERSION IN WORD IS AVAILABLE ON THE SCHOOLPHYSICS USB

© Keith Gibbs 2020