Problem 2-43 - Free fall - Part 6 (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

\(\mathrm{t = 4.2\; s \\ a_y = -9.8\; m/s^2 \\ v_y = -v_{0_y}}\)

Correct!

\(\mathrm{v_y = v_{0_y} + a_yt}\)
We can also use
\(\mathrm{y = y_0 + v_{0_y}t + ½ \;a_yt^2}\) with \(\mathrm{y_0 = 0}\)
If \(\mathrm{y_0 = 0}\) it reduces to \(\mathrm{v_y = v_{0_y} + a_yt}\)
Substituting \(\mathrm{-v_{0_y} = v_{0_y} + (-9.8)(4.2)}\)
\(\mathrm{2\; v_{0_y} = 41.16}\)
\(\mathrm{v_{0_y} = 21 \;m/s}\)
This is the answer to part (a)

If the total time to rise and fall is \(4.2\; s\), what is the time to rise to the maximum height?
 
(A)   \(4.2\; s\)

(B)   \(2.1 \;s\)

(C)   \(8.4 \;s\)