You throw a stone straight down from the top of a tall tower. It leaves your hand moving at 8.00 m/s, Air resistance can be neglected. Take

Question

You throw a stone straight down from the top of a tall tower. It leaves your hand moving at 8.00 m/s, Air resistance can be neglected. Take the positive -direction to be upward, and choose y 0 to be the point where the stone leaves your hand. Find the stone’s position 1.50s after it leaves your hand.
Express your answer with the appropriate units.
Find the y-component of the stone’s velocity 1.50 s after it leaves your hand. Express your answer with t0he appropriate units.

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Dulcie 4 years 2021-07-19T12:09:02+00:00 1 Answers 28 views 0

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  1. thienthi
    0
    2021-07-19T12:10:24+00:00
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    Answer:

    The velocity after 1.5 s is 22.7 m/s downwards.

    Explanation:

    Initial velocity = – 8 m/s

    acceleration, a = – 9.8 m/s2

    time, t = 1.5 s

    Use first equation of motion

    v = u + at

    v = – 8 – 9.8 x 1.5

    v = – 8 – 14.7

    v = – 22.7 m/s  

    Thus, the velocity after 1.5 s is 22.7 m/s downwards.

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