A ball thrown vertically is caught by the thrower after 5.1s. Find the maximum height the ball reaches.

A ball thrown vertically is caught by the thrower after 5.1s. Find the maximum height the ball reaches.

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  1. Answer:

    h = 31.9 m

    Explanation:

    Since, the ball took 5.1 s in the air. Hence, the time to reach maximum height will be equal to the half of this value:

    t = 5.1 s /2 = 2.55 s

    Now, we use 1st equation of motion between the time of throwing and the time of reaching maximum height:

    Vf = Vi + gt

    where,

    Vf = Final Velocity = 0 m/s (since, ball momentarily stops at highest point)

    Vi = Initial Velocity = ?

    g = – 9.8 m/s² (negative sign for upward motion)

    Therefore,

    0 m/s = Vi + (-9.8 m/s²)(2.55 s)

    Vi = 24.99 m/s

    Now, we use second equation of motion for height (h):

    h = Vi t + (0.5)gt²

    h = (24.99 m/s)(2.55 s) + (0.5)(-9.8 m/s²)(2.55 s)²

    h = 63.7 m – 31.8 m

    h = 31.9 m

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