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28 June, 10:28

A projectile is fired at an upward angle of 29.7° from the top of a 108-m-high cliff with a speed of 130-m/s. What will be its speed (in m/s) when it strikes the ground below?

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  1. 28 June, 10:51
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    79.2 m/s

    Explanation:

    θ = angle at which projectile is launched = 29.7 deg

    a = initial speed of launch = 130 m/s

    Consider the motion along the vertical direction

    v₀ = initial velocity along the vertical direction = a Sinθ = 130 Sin29.7 = 64.4 m/s

    y = vertical displacement = - 108 m

    a = acceleration = - 9.8 m/s²

    v = final speed as it strikes the ground

    Using the kinematics equation

    v² = v₀² + 2 a y

    v² = 64.4² + 2 (-9.8) (-108)

    v = 79.2 m/s
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