Chapter 2, Problem 37P

To determine

The speed of ball while throwing and the height reached by the ball.

Answer to Problem 37P

  $14.7\frac{\text{m}}{\text{s}}$ and $11.025\text{m}$

Explanation of Solution

Given:

Initial velocity of the ball as it is thrown $=v_i$

Time taken by ball to travel in air $=t=3\text{s}$

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

Initial position of ball when it was thrown $=y_i=y$

Final position of ball when it is caught at same height $=y_f=y$

Final velocity of the ball at the maximum height $=v_f=0\frac{\text{m}}{\text{s}}$

Height gained by the ball $=h$

Final position of ball at maximum height $=y_{fmax}=h$

Formula Used:

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

  $y_f=y_i+v_{i}t+(0.5)a{t}^2$

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

  $v_f^2=v_i^2+2a(y_f-y_i)$

Where,

  $v_i=$ Initial velocity

  $v_f=$ Final velocity

  $a=$ Acceleration

  $y_f=$ Final position

  $y_i=$ Initial position

  $t=$ Time

Calculation:

Consider the motion of the ball in air.

Substituting the values

  $\begin{array}{l}y=y+v_i(3)+(0.5)(-9.8){(}^3\\ y-y=v_i(3)-44.1\\ v_i(3)=44.1\\ v_i=14.7\frac{\text{m}}{\text{s}}\end{array}$

Consider the motion of the ball from launching point to the maximum height.

Use the kinematics equation

  $v_f^2=v_i^2+2a(y_{f\max }-y_i)$

Substituting the values

  $\begin{array}{l}{0}^2={14.7}^2+2(-9.8)(h-0)\\ h=11.025\text{m}\end{array}$

Conclusion:

Therefore, initial speed and height gained comes out to be $14.7\frac{\text{m}}{\text{s}}$ and $11.025\text{m}$