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?

0.938 seconds

Explanation:

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

where s = distance moved

u = initial speed = 19.2 m/s

t = time taken

g = acceleration due to gravity = 9.8

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

=> ________(1)

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

u = 0 m/s

At the point where both balls are level:

s = 18 - x

=>

=> __________(2)

Equating both (1) and (2):

They will be level after 0.938 seconds

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