A 42.0 kg projectile is launched horizontally from the top of a cliff, 50 meters above the ground. The projectile’s initial speed is 30 m/s.

Question

A 42.0 kg projectile is launched horizontally from the top of a cliff, 50 meters above the ground. The projectile’s initial speed is 30 m/s. What is the velocity right before it hits the ground

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Thanh Thu 4 years 2021-08-13T14:24:20+00:00 1 Answers 7 views 0

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  1. danthu
    0
    2021-08-13T14:25:41+00:00
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    Answer:

    Velocity of the projectile just before it will hit the surface is 43.4 m/s

    Explanation:

    In x direction there is no acceleration

    so horizontal velocity of the projectile is always constant

    So we have

    v_x = 30 m/s

    now in Y direction its motion is free fall motion under gravity

    so we have

    v_y^2 = 0^2 + 2(9.81)(50)

    v_y = 31.3 m/s

    so we have

    v = \sqrt{v_y^2 + v_x^2}

    v = \sqrt{30^2 + 31.3^2}

    v = 43.4 m/s

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