A ball is thrown in the air vertically from ground level with initial velocity 2

Question

A ball is thrown in the air vertically from ground level with initial velocity 288 ft/s. Find the average height of the ball over the time interval extending from the time of the ball?s release to its return to ground level. The height at time t is h(t) = 288t - 16t^2. Your answer must include units A = 864m (If you are unable to figure this problem out. you will receive a hint after 3 tries) HINT: The height of the ball can be represented by the graph above. In order to find the average height (which you have the formula for), what other information do you need. and how can you get it?

Explanation / Answer

h(t)=288t -16t^2

initially and finally 288t -16t^2 =0

==>16t(18 -t) =0

==>t=0,t=18

average height =(1/(18-0))integral[0 to 18] 288t -16t^2 dt

=(1/(18-0)) [0 to 18] 144t^2 -(16/3)t^3 +c

=(1/18)  144*18^2 -(16/3)*18^3 -0

=144*18 -(16/3)*18^2

=864ft

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