Question

A ball thrown in the air vertically from ground level with initial velocity 37 m/s has height h(t) = 37t − 4.9t2 m at time t (in seconds). Find the average height over the time interval extending from the ball's release to its return to ground level. (Round your answer to two decimal places.)

Answer 1

Answer:

46.57

Step-by-step explanation:

Given that:

h(t) = 37t - 4.9t²

When the ball reaches the ground, then the height is zero;

So, we equate h(t) to zero and solve for t

37t - 4.9t² = 0

t = 0, 7.55

Let the Initial time be  a = t = 0

The time the ball reaches the ground be b = 7.55

So, the average height can now be calculated as:

Average height = $\frac{1}{b-a}\int\limits^b_a {h(t)} \, dt$

= $\frac{1}{7.55-0}\int\limits^{7.55}_0 (37t-4.9^2) dt$

= $\frac{1}{7.55-0}\ (\frac{37(7.55)^2}{2} - \frac {4.9(7.55)^3}{3})$

= $\frac{1}{7.55-0}\ (351.6104208)$

= 46.57091665

≅ 46.57  to (2 decimal places)