Question

A ball is thrown in the air vertically from ground level with initial velocity 288ft/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.
Recall that the height at time t is h(t)=288t−16t²
Your answer must include units. A=

Answer 1

The average height of the ball over the time interval from its release to its return to ground level is 0 ft.

Step-by-step explanation:

To find the average height of the ball over the time interval extending from its release to its return to ground level, we need to calculate the total time it takes for the ball to reach the ground.

We can do this by setting the height function equal to zero and solving for t:

0 = 288t - 16t²

By factoring out a t from the equation, we get:

0 = t(288 - 16t)

From this equation, we can see that t = 0 or t = 18. Now, since we're looking for the average height over the time interval from the ball's release to its return to ground level, we only need to consider the positive value of t.

Therefore, the ball takes 18 seconds to reach the ground. Now, we can substitute this value into the height function to find the average height:

Average height = h(18) = 288 * 18 - 16 * 18² = 5184 - 5184 = 0 ft

So, the average height of the ball over the time interval from its release to its return to ground level is 0 ft.