Chapter 2, Problem 41P

To determine

To show: The speed of ball while coming back is same as the speed of the ball while going up.

Explanation of Solution

Given:

Initial velocity of the ball $=v_i=v_o$

Final velocity of the ball $=v_f$

Acceleration experienced by the ball = acceleration due to gravity $=-9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Initial position of the ball $=y_i=0\text{m}$

Final position of the ball $=y_f=0\text{m}$

Formula Used:

Kinematics equation that relates final velocity, initial velocity, acceleration, initial and final position of an object is given as

  $v_f^2=v_i^2+2a(y_f-y_i)$

Calculation:

Substituting the values,

  $\begin{array}{l}{v}_f^2=v_o^2+2(-9.8)(0-0)\\ v_f^2=v_o^2+0\\ v_f^2=v_o^2\\ v_f=v_o\end{array}$

Conclusion:

Therefore, final speed of the ball is same as the initial speed.