Find Its Average Velocity From The Time It Is Thrown (t=0) To The Time It Reaches Its Maximum Height. Can You Help?

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Oddman Profile
Oddman answered
The acceleration due to gravity is a constant, so the decrease in velocity with respect to time is linear. Thus, the average velocity from the time the ball is thrown until the velocity reaches zero is (1/2)*(80 ft/s) = 40 ft/s.

Consider a plot of velocity vs. Time. The (negative) slope is the acceleration due to gravity. The (triangular) area under the velocity curve is the distance traveled (it has units of velocity times time), which is
(distance traveled) = (1/2)(initial velocity)(time to zero velocity).

Thus the average velocity is (distance traveled)/(time to zero velocity), the number given above.

Richard Enison Profile
Richard Enison answered
The instantaneous velocity at time t is the derivative of the height y, or 80 - 32t. This is 0 when t = 80/32 = 5/2 = 2.5 sec. The height at that time is y = 80 (2.5) - 16 (2.5)2 = 200 - 100 = 100 ft. The average velocity is therefore 100/2.5 = 40 ft/sec.

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