Question

Ill do the rest.
A ball was thrown up at 25.0 m/s. It was released when it was 2.0 m above the ground.
a) For how long did the ball rise?
b) at the top of its trajectory, how far above the ground was it?
c) what was the ball's speed, direction and height 2.0 s after release?
d) what was its speed, direction and height after 4.0 s after release?

Answer 1

v = Vi - 9.8 t
when is v zero (stopped at top) ?
0 = 25 - 9.8 t
solve for t

Answer 2

Thank you very much. :)

Answer 3

You are welcome :)