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A ball is thrown horizontally from the top of a building 109 m high. The ball strikes the ground at a point 75 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground? (note: speed is magnitude of velocity at that point).

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Answer:

The speed of the ball just before it strikes the ground is 15.90 m/s.

Step-by-step explanation:

Given that,

Height = 109 m

Distance = 75 m

We need to calculate the time

Using formula for horizontal projectile


t=\sqrt{(2h)/(g)}

Where, h = height

Put the value into the formula


t=\sqrt{(2*109)/(9.8)}


t=4.716\ sec

We need to calculate the speed of the ball just before it strikes the ground

Using formula of speed


v =(d)/(t)

Where, v = speed

d = distance

t = time

Put the value into the formula


v=(75)/(4.716)


v=15.90\ m/s

Hence, The speed of the ball just before it strikes the ground is 15.90 m/s.

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