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10 July, 18:07

You throw a rock straight up and find that it returns to your hand 3.40 s after it left your hand. Neglect air resistance. What was the maximum height above your hand that the rock reached?

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  1. 10 July, 19:59
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    The maximum height of the rock is 14.2 m

    Explanation:

    The equations that describe the height and velocity of the rock are the following:

    y = y0 + v0 · t + 1/2 · g · t²

    v = v0 + g · t

    Where:

    y = height of the object at time t

    y0 = initial height

    t = time

    g = acceleration due to gravity (-9.8 m/s² if upward is positive)

    v = velocity of the object at time t

    We know that at t = 3.40 s, the rock is in your hand again. Then, if we place the origin of the frame of reference at your hand, the position of the rock at 3.40 s is 0 m. Using the equation of the position, we can calculate the initial velocity that we will need to obtain the max-height.

    y = y0 + v0 · t + 1/2 · g · t²

    0 = v0 · 3.40 s - 1/2 · 9.8 m/s² · (3.40 s) ²

    (1/2 · 9.8 m/s² · (3.40 s) ²) / 3.40 s = v0

    v0 = 16.7 m/s

    At max-height, the velocity of the rock is 0. Then, using the equation of velocity we can calculate the time it takes the rock to reach the max-height. With that time, we can calculate the maximum height.

    v = v0 + g · t (at max-height, v = 0)

    0 = 16.7 m/s - 9.8 m/s² · t

    - 16.7 m/s / - 9.8 m/s² = t

    t = 1.70 s

    Now, using this time in the equation of height:

    y = y0 + v0 · t + 1/2 · g · t²

    y = 0 m + 16.7 m/s · 1.70 s - 1/2 · 9.8 m/s² · (1.70 s) ²

    y = 14.2 m

    The maximum height of the rock is 14.2 m
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