A ball is thrown straight up from the top of a building 144 ft tall with an initial velocity of 48 ft per second. The height s(t) (in

Question

A ball is thrown straight up from the top of a building 144 ft tall with an initial velocity of 48 ft per second. The height
s(t) (in feet)
of the ball from the ground, at time t (in seconds), is given by
s(t) = 144 + 48t − 16t^2.
Find the maximum height attained by the ball.

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Thu Cúc 2 months 2021-08-22T11:03:00+00:00 1 Answers 6 views 0

Answers ( )

    0
    2021-08-22T11:04:29+00:00

    Answer:

    180feet

    Step-by-step explanation:

    Given the function that models the height of a ball

    s(t) = 144 + 48t − 16t^2.

    At maximum height, the velocity is zero ie. ds/dt  = 0

    ds/dt = 48 – 32t

    0 = 48 – 32t

    48 = 32t

    t = 48/32

    t = 1.5secs

    Get the maximum height

    s(t) = 144 + 48t − 16t^2.

    s(1.5) = 144 + 48(1.5)-16(1.5)^2

    s()1.5) = 144 + 72 – 16(2.25)

    s(1.5) = 144 + 72-36

    s(1.5) = 180 feet

    Hence the maximum height attained by the ball is 180feet

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