A ball is thrown into the air and its position is given by h ( t ) = − 6.3 t 2 + 53 t + 24 where h is the height of the ball in meters t sec

Question

A ball is thrown into the air and its position is given by h ( t ) = − 6.3 t 2 + 53 t + 24 where h is the height of the ball in meters t seconds after it has been thrown. Find the maximum height reached by the ball and the time at which that happens.

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Raelynn 3 weeks 2021-10-01T11:40:58+00:00 1 Answer 0

Answers ( )

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    2021-10-01T11:42:34+00:00

    Answer:

    h'(t) = -12.6 t +53

    Now we can set up the derivate equal to 0 and we have:

     -12.6 t +53 = 0

    And solving for t we got:

     t = \frac{53}{12.6}= 4.206

    For the second derivate respect the time we got:

     h''(t) = -12.6<0

    So then we can conclude that t = 4.206 is a maximum for the function.

    And the corresponding height would be:

     h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft

    So the maximum occurs at t = 4.206 s and with a height of 135.468 m

    Step-by-step explanation:

    For this case we have the following function:

     h(t) = -6.3t^2 +53 t+24

    In order to maximize this function we need to take the first derivate respect the time and we have:

    h'(t) = -12.6 t +53

    Now we can set up the derivate equal to 0 and we have:

     -12.6 t +53 = 0

    And solving for t we got:

     t = \frac{53}{12.6}= 4.206

    For the second derivate respect the time we got:

     h''(t) = -12.6<0

    So then we can conclude that t = 4.206 is a maximum for the function.

    And the corresponding height would be:

     h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft

    So the maximum occurs at t = 4.206 s and with a height of 135.468 m

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