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11 February, 07:44

Two physics students shoot a water-bottle rocket from a 1 m tall stand. They calculate that rocket will travel with the given formula, f (x) = -16t 2+64t+1 where f is its height in feet above ground t seconds after being released. After how many seconds will the rocket reach its maximum height? What is that height?

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  1. 11 February, 07:48
    0
    2 secs; 65 feet

    Explanation:

    The function guiding the water bottle is given as:

    f (x) = - 16t² + 64t + 1

    The bottle will reach maximum height when velocity, df/dt (velocity is the first derivative of distance) = 0.

    Therefore:

    df/dt = 0 = - 32t + 64

    => 32t = 64

    t = 64/32 = 2 seconds

    This is the time it will take to reach the maximum height.

    To find this height, we insert t = 2 into the function f (x):

    f = - 16 (2) ² + 64 (2) + 1

    f = - (16 * 4) + 128 + 1

    f = - 64 + 128 + 1

    f = 65 ft

    Its maximum height is 65 ft.
  2. 11 February, 08:00
    0
    It takes the water bottle rocket 2 s to reach maximum height.

    The maximum height reached is 65 m

    Explanation:

    The object would reach maximum height when its velocity is zero. That is f' (x) = 0. So, we differentiate f (x) to get

    df (x) / dx = d (-16t² + 64t + 1) / dt = - 32t + 64 = 0

    -32t + 64 = 0

    -32t = - 64

    t = - 64/-32 = 2 s

    So it takes the water bottle rocket 2 s to reach maximum height.

    To find this height, we substitute t = 2 into f (x). So,

    f (x) = - 16t² + 64t + 1

    f (2) = - 16 (2) ² + 64 (2) + 1 = - 16 (4) + 128 + 1 = - 64 + 128 + 1 = 65 m

    So the maximum height reached is 65 m.
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