Brandon is standing on top of a 50.0m tall building. He happens to notice a bouncy ball in his pocket and throws it up with a velocity of 20.0m/s. What is the average velocity of the ball as it was...

Brandon is standing on top of a 50.0m tall building. He happens to notice a bouncy ball in his pocket and throws it up with a velocity of 20.0m/s. What is the average velocity of the ball as it was going up?

Expert Answers
gsenviro eNotes educator| Certified Educator

The average velocity is given as the ratio of change in position over a given time interval. Here, no time interval is given. When the ball is thrown up, it will decelerate and ultimately come to rest and then start falling back to the ground.

Time taken to reach the highest point in its trajectory, t:

`t = (v-u)/a = (0-20)/(-9.81) sec = 2.04 sec` 

The maximum height gained is given as, H:

`H = (v^2-u^2)/(2a) = (0^2-20^2)/(-2xx9.81) = 20.39 m`

Thus the average velocity of the upward travel, Vavg = H/t

`V_(avg) = H/t = 20.39/2.04 m/s = 9.995 m/s or 10.0 m/s`

Thus, the average velocity of the ball as it was going up was 10 m/s.

Hope this helps.

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