Problem 2-43 - Free fall - Part 8 (c)

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

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

Time to rise to maximum height \(\mathrm{= 2.1\; s}\)

\(\mathrm{v_{0_y} = 21\; m/s}\)

Correct!

The velocity at maximum height is zero.
 

Using \(\mathrm{v_y{^2} - v0_y{^2} = 2a_yy}\)

\(\mathrm{0^2 = (21)^2 + 2(-9.8)y}\)

\(\mathrm{y = 441/19.6 = 22\; m}\) as before

You have completed this problem.