Chapter 2, Problem 83GP

To determine

The speed with which a friend throw the ball

Answer to Problem 83GP

  $11.8\frac{\text{m}}{\text{s}}$

Explanation of Solution

Given:

Initial velocity of the ball dropped from top of cliff $=v_{1o}=0\frac{\text{m}}{\text{s}}$

Height of the cliff $=h$

Initial position of the ball at the top of the cliff $=y_{1o}=h$

Final position of the ball at the bottom of the cliff $=y_{1f}=0\text{m}$

Time of travel for ball dropped $=t_1=1.2\sec$

Acceleration of the ball dropped $=a=-9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Initial velocity of the ball thrown up $=v_{2o}$

Final velocity of the ball thrown up $=v_{2f}=0\frac{\text{m}}{\text{s}}$

Initial position of the ball thrown up $=y_{2o}=0\text{m}$

Final position of the ball thrown up $=y_{2f}=h$

Acceleration of the ball thrown up $=a=-9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Formula Used:

Position of an object at any time in accelerated motion is given as

  $y_f=y_o+v_{o}t+(0.5)a{t}^2$

Acceleration is given as

  $a=\frac{v_{2f}^2-v_{2o}^2}{2(y_f-y_o)}$

Calculation:

Consider the motion of the ball dropped from the top of the cliff.

Final position of the ball dropped is given as

  $y_{1f}=y_{1o}+v_{1o}{t}_1+(0.5)a{t}_1{}^2$

Substituting the values

  $\begin{array}{l}0=h+(0)(1.2)+(0.5)(-9.8){(}^{1.2}\\ h=7.056\text{m}\end{array}$

Consider the motion of the ball thrown up by the friend.

Acceleration of the ball thrown up is given as

  $a=\frac{v_{2f}^2-v_{2o}^2}{2(y_f-y_o)}$

Substituting the values

  $\begin{array}{l}-9.8=\frac{0^2-v_{2o}^2}{2(h-0)}\\ -9.8=\frac{0^2-v_{2o}^2}{2(7.056-0)}\\ v_{2o}=11.8\frac{\text{m}}{\text{s}}\end{array}$

Conclusion:

Therefore, the speed of the ball thrown by friend comes out to be $11.8\frac{\text{m}}{\text{s}}$