A ball is thrown straight up with initial velocity of 64 ft./s. The height of the ball t seconds after it is thrown is given by the formula

Question

A ball is thrown straight up with initial velocity of 64 ft./s. The height of the ball t seconds after it is thrown is given by the formula s(t)=64t-16t^2

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Acacia 4 years 2021-08-21T15:24:17+00:00 1 Answers 33 views 0

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  1. binhan
    0
    2021-08-21T15:26:04+00:00
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    Answer:

    64 feet

    Step-by-step explanation:

    Complete question

    A ball is thrown straight up with initial velocity of 64 ft./s. The height of the ball t seconds after it is thrown is given by the formula s(t)=64t-16t^2. Find the maximum height reached by the ball.

    At maximum height, the velocity of the ball is zero, i.e ds/dt  =0

    ds/dt = 64 – 32t

    64 – 32t = 0

    32t = 64

    t = 64/32

    t = 2secs

    Substitute t = 2secs into the given equation;

    Recall that s(t)=64t-16t^2

    s(2) = 64(2) – 16(2)^2

    s(2) = 128 – 16(4)

    s(2) = 128 – 64

    s(2) = 64 ft

    Hence the maximum height reached by the ball is 64 feet

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