Question

A ballplayer catches a ball 3.0s after throwing it vertically upward. With what speed did he throw it, and what height did it reach?

Answer: 15 m/s and 11m

Answer 1

Are you asking for some work shown, or what?

Answer 2

First note that the time it took for it to reach the max height will be half the time it took for the ballplayer to catch it and at the max height, the y velocity will be 0.

To find the initial velocity you can use the kinematic equation:
\[V_f-Vi=at\]\[0m/s-Vi=(-9.8m/s^2)(1.5s)\]\[V_i=14.7m/s\]To find the max height, you can use this kinematic equation:
\[d=V_it+\frac{1}{2}at^2\]\[d=(14.7m/s)(1.5s)+\frac{1}{2}(-9.8m/s^2)(1.5s)^2=11.025m\]Once you round those you'll get the answers you posted.

Answer 3

Thank you so much

Answer 4

yw :)