## You toss a 0.150-kg baseball straight upward so that it leaves your hand moving at 20.0 m/s. The ball reaches a maximum height y2. What is t

Question

You toss a 0.150-kg baseball straight upward so that it leaves your hand moving at 20.0 m/s. The ball reaches a maximum height y2. What is the speed of the ball when it is at a height of y2/2? Ignore air resistance.

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Physics
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2021-07-22T19:26:19+00:00
2021-07-22T19:26:19+00:00 1 Answers
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## Answers ( )

Answer:14.14 m/s

Explanation:mass of ball, m 0.150 kg

initial velocity, u = 20 m/s

g = 10 m/s²

The final velocity at the height is zero. Total height is y2.

Use third equation of motion,

v² = u² – 2 gh0 = 20 x 20 – 2 x 10 x y2

y2 = 20 m

Let v is the velocity at height y2/2 .

v² = u² – 2 ghv² = 20 x 20 – 2 x 10 x 10

v² = 400 – 200

v = 14.14 m/s

Thus, the velocity at half of the height is 14.14 m/s.