A ball is launched straight up in the air from a height of 6 feet. Its velocity​ (feet/second) t seconds after launch is given by f(t)= -34t

Question

A ball is launched straight up in the air from a height of 6 feet. Its velocity​ (feet/second) t seconds after launch is given by f(t)= -34t+291.
Between 1 second and 8 seconds, the​ ball’s height changed by … feet.
​(Round answer to the nearest​ tenth.)

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Thu Thủy 2 weeks 2021-09-02T05:39:59+00:00 1 Answers 0 views 0

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    2021-09-02T05:41:50+00:00

    Answer:

    966 feet

    Step-by-step explanation:

    First, we can see that we have the velocity function given, a base value for the height, and need to figure out the change in height. We also know that velocity is the derivative of position/height. Thus, we can find the integral of the velocity to find an equation for the height.

    \int\limits^1_0 {-34t+291} \, dt\\

    Use the exponent rule to turn -34t into -17t² and 291 into 291t to get our result as

    -17t²+291t + C = height (h)

    When t=0, we know that our height (h) is 6, so C = 6, making our equation

    -17t²+291t + 6

    To find the change between 1 second and 8 seconds, we can plug 1 and 8 in for t, and find the difference between those values, which is

    (-17(8)²+291(8) + 6) – ( -17(1)²+291(1) + 6 )

    = 1246 – 280

    = 966

    Note that we did final value (t = 8) – initial value (t=1), not the other way around

    The change in height, between 1 second and 8 seconds, is therefore 966 feet

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