Suppose that a baseball is tossed up into the air at an initial velocity 1818​​\text{ m/s} m/s​. The height of the baseball at time tt​ in s

Question

Suppose that a baseball is tossed up into the air at an initial velocity 1818​​\text{ m/s} m/s​. The height of the baseball at time tt​ in seconds is given by h\mathopen{}\mathclose{\left(t\right)}=18 t-4.9 t^{2}h(t)=18t−4.9t 2 ​​​​​ (in meters). a) What is the average velocity for \left[1, 1.5\right][1,1.5]​?

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Latifah 11 hours 2021-07-21T21:07:32+00:00 1 Answers 0 views 0

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    2021-07-21T21:09:06+00:00

    Answer:

    13.1 m/s

    Explanation:

    Given that a baseball is tossed up into the air at an initial velocity 18 m/s​. The height of the baseball at time t​ in seconds is given by h(t) = 18t−4.9t 2 ​​​​​ (in meters).

    a) What is the average velocity for [1,1.5]​?

    To calculate the velocity travelled by the ball, differentiate the function.

    dh/dt = 18 – 9.8t

    Substitute t for 1 in the above Differential function

    dh/dt = 18 – 9.8 (1)

    But dh/dt = velocity

    V = 18 – 9.8

    V = 8.2 m/s

    Average velocity = ( U + V ) / 2

    Average velocity = (18 + 8.2)/2

    Average velocity = 26.2/2

    Average velocity = 13.1 m/s

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