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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Thu Nguyệt 3 days 2021-07-22T19:26:19+00:00 1 Answers 2 views 0

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    2021-07-22T19:27:32+00:00

    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 gh

    0 = 20 x 20 – 2 x 10 x y2

    y2 = 20 m

    Let v is the velocity at height y2/2 .

    v² = u² – 2 gh

    v² = 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.

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