Chapter 2, Problem 82GP

To determine

The height of ball of person B in comparison to person J.

Answer to Problem 82GP

2.25 times

Explanation of Solution

Given:

Initial velocity of person J ball $=v_{jo}=v_o$

Initial velocity of person B ball $=v_{bo}=1.5v_o$

Final velocity of each ball $=v_f=0\frac{\text{m}}{\text{s}}$

Displacement of person J ball $=\Delta y_j$

Displacement of person B ball $=\Delta y_b$

Average acceleration of the ball $=a$

Formula Used:

Displacement of the ball is given as

  $\Delta y=\frac{v_f^2-v_o^2}{2a}$

Calculation:

Displacement of the ball when person J throws it is given as

  $\begin{array}{l}\Delta y_j=\frac{v_f^2-v_{jo}^2}{2a}\\ \Delta y_j=\frac{0^2-v_o^2}{2a}\\ \Delta y_j=\frac{-v_o^2}{2a}\left(1\right)\end{array}$

Displacement of the ball when bill throws it is given as

  $\begin{array}{l}\Delta y_b=\frac{v_f^2-v_{bo}^2}{2a}\\ \Delta y_b=\frac{0^2-{(1.5v_o)}^2}{2a}\\ \Delta y_b=\frac{-2.25v_o^2}{2a}\left(2\right)\end{array}$

Divide equation (2) by equation (1).

  $\begin{array}{l}\frac{\Delta y_b}{\Delta y_j}=\frac{\frac{-2.25v_o^2}{2a}}{\frac{-v_o^2}{2a}}\\ \frac{\Delta y_b}{\Delta y_j}=2.25\\ \Delta y_b=(2.25)\Delta y_j\end{array}$

Conclusion:

Therefore, person B ball goes $2.25$ higher.