Problem 2-43 - Free fall - Part 7 (a)
A baseball pitcher throws a ball vertically straight upward and catches it \(4.2 \;s\) later. (a) With what velocity did the ball leave the pitcher's hand? (b) What was the maximum height reached by the ball?
Accumulated solution
\(\mathrm{t = 4.2\; s \\ a_y = -9.8\; m/s^2 \\ v_y = -v_{0_y}}\)
No. The time to rise to maximum height is \(½\) the total time or \(4.2/2 = 2.1\; s\)