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3 August, 05:41

A rock is thrown from the top of a 20-m building at an angle of 53° above the horizontal. If the horizontal range of the throw is equal to the height of the building, with what speed was the rock thrown? What is the velocity of the rock just before it strikes the ground?

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  1. 3 August, 06:55
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    28.5 m/s

    18.22 m/s

    Explanation:

    h = 20 m, R = 20 m, theta = 53 degree

    Let the speed of throwing is u and the speed with which it strikes the ground is v.

    Horizontal distance, R = horizontal velocity x time

    Let t be the time taken

    20 = u Cos 53 x t

    u t = 20/0.6 = 33.33 ... (1)

    Now use second equation of motion in vertical direction

    h = u Sin 53 t - 1/2 g t^2

    20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)

    t = 1.17 s

    Put in equation (1)

    u = 33.33 / 1.17 = 28.5 m/s

    Let v be the velocity just before striking the ground

    vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s

    vy = uSin 53 - 9.8 x 1.17

    vy = 28.5 x 0.8 - 16.66

    vy = 6.14 m/s

    v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2

    v = 18.22 m/s
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