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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.)

1 Answer

7 votes

Answer:

46.57

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 =
(1)/(b-a)\int\limits^b_a {h(t)} \, dt

=
(1)/(7.55-0)\int\limits^(7.55)_0 (37t-4.9^2) dt

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

=
(1)/(7.55-0)\ (351.6104208)

= 46.57091665

≅ 46.57 to (2 decimal places)

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