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26 September, 09:23

A bowling ball is launched off the top of a 240-foot tall building. The height of the bowling ball above the ground t seconds after being launched is s (t) = -16t2+32t+240s (t) = -16t2+32t+240 feet above the ground. What is the velocity of the ball as it hits the ground?

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  1. 26 September, 09:27
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    When the ball hits the ground, its velocity is - 128 ft/s.

    Explanation:

    Hi there!

    First, let's find the time it takes the ball to reach the ground (the value of t for which s (t) = 0):

    s (t) = - 16t² + 32t + 240

    0 = - 16t² + 32t + 240

    Solving the quadratic equation with the quadratic formula:

    t = 5.0 s (the other solution of the equation is rejected because it is negative).

    Now, we have to find the velocity of the ball at t = 5.0 s.

    The velocity of the ball is the change of height over time (the derivative of s (t)):

    v = ds/dt = s' (t) = - 32t + 32

    at t = 5.0 s:

    s' (5.0) = - 32 (5.0) + 32 = - 128 ft/s

    When the ball hits the ground, its velocity is - 128 ft/s.
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