Question

A ball is thrown upward from the ground with an initial speed of 19.2 m/s; at the same instant, another ball is dropped from a building 18 m high. After how long will the balls be at the same height above the ground?

Answer 1

Answer:

0.938 seconds

Explanation:

For the ball thrown upwards, we use the formula below to solve it:

$s = ut - \frac{1}{2}gt^2$

where s = distance moved

u = initial speed = 19.2 m/s

t = time taken

g = acceleration due to gravity = 9.8 $m/s^2$

Let x be the height at which both balls are level, this means that:

=> $x = 19.2t - 4.9t^2$________(1)

For the ball dropped downwards, we use the formula below:

$s = ut + \frac{1}{2}gt^2$

u = 0 m/s

At the point where both balls are level:

s = 18 - x

=> $18 - x = 0 + 4.9t^2$

=> $x = 18 - 4.9t^2$__________(2)

Equating both (1) and (2):

$19.2t - 4.9t^2 = 18 - 4.9t^2\\\\=> 19.2t = 18\\\\t = 18/19.2 = 0.938 secs$

They will be level after 0.938 seconds