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9 January, 14:22

A rock is thrown off of a 100 foot cliff with an upward velocity of 45 m/s. As a result its height after t seconds is given by the formula?

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  1. 9 January, 14:29
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    Q: A rock is thrown off of a 100 foot cliff with an upward velocity of 45 m/s. As a result its height after t seconds is given by the formula:

    h (t) = 100+45t-4.9t2

    (a) What is its height after 3 seconds?

    (b) What is its velocity after 3 seconds?

    Answer:

    (a) 190.9 m.

    (b) 15.6 m/s upward

    Explanation:

    Given:

    h (t) = 100 + 45t - 4.9t²

    The height after 3 seconds,

    t = 3 s

    Substitute the value of t in to the equation above.

    h (3) = 100+45 (3) - 4.9 (3) ²

    h (3) = 100+135-44.1

    h (3) = 190.9 m

    Therefore the height after 3 seconds = 190.9 m.

    (b) Velocity after 3 seconds

    The velocity is obtained by differentiating h (t) with respect to time

    v = dh (t) / dt

    dh (t) / dt = 45-9.8t

    v = 45 - 9.8t ... Equation 1

    t = 3 s.

    Substitute the value of t into the equation above,

    v = 45 - 9.8 (3)

    v = 45 - 29.4

    v = 15.6 m/s

    Thus the velocity after 3 seconds = 15.6 m/s upward
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