Problem 2-43 - Free fall - Part 7 (b)

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 \(= 2.1 \;s\)

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

Correct!

The time to rise to maximum height is \(½ (4.2) = 2.1\; s\)
 
Using \(\mathrm{y = y_0 + v_{0_y}t + ½ a_yt^2}\) with \(\mathrm{y_0 = 0}\)
 
we have
 
\(\mathrm{y = 0 + 21(2.1) + ½ (9.8)(2.1)^2= 22\; m}\)

This is the answer to part (b)

Or using \(\mathrm{v_y{^2} - v_{0_y}{^2} = 2a_yy}\)
 

what is the value of \(v_y\) at the maximum height? 
 

(A)   \(21\; m/s\)

(B)   \(-21\; m/s\)

(C)   \(0\)