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PilotLPTM [1.2K]
3 years ago
9

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 b

uilding 18 m high. After how long will the balls be at the same height above the ground?
Physics
1 answer:
aleksley [76]3 years ago
4 0

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

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