Chapter 2, Problem 36P

(a)

To determine

The height gained by the baseball

Answer to Problem 36P

24.7 m.

Explanation of Solution

Given:

Initial velocity of the baseball $=v_i=22\frac{\text{m}}{\text{s}}$

Final velocity of the baseball at the highest point $=v_f=0\frac{\text{m}}{\text{s}}$

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

Height gained by the ball $=h$

Displacement of the ball $=\Delta y=h$

Formula Used:

Displacement of an object in uniform accelerated motion is given as

  $\begin{array}{l}\Delta y=\frac{v_f^2-v_i^2}{2a}\\ \text{Where }{v}_f\text{ and }{v}_i\text{ are final and initial velocity of the object and }a\text{ is the acceleration}\end{array}$

Calculation:

Consider the motion of the baseball in vertical direction.

Displacement of the baseball is given as

  $\Delta y=\frac{v_f^2-v_i^2}{2a}$

Substituting the values

  $\begin{array}{c}\Delta y=\frac{0^2-{22}^2}{2(-9.8)}\\ =24.69\text{m}\\ \Delta y\approx 24.7\text{m}\end{array}$

Conclusion:

Therefore, height gained is 24.7 m.

(b)

To determine

The time spent by the baseball in air

Answer to Problem 36P

4.5 seconds.

Explanation of Solution

Given:

Initial velocity of the baseball $=v_i=22\frac{\text{m}}{\text{s}}$

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

Time of travel $=t$

Displacement of the ball as it comes back to its starting location $=\Delta y=0\text{m}$

Formula Used:

Displacement of an object in uniform accelerated motion is given as

  $\begin{array}{l}\Delta y=v_{i}t+(0.5)a{t}^2\\ \text{where }{v}_i\text{= initial velocity}\\ \Delta y=\text{displacement}\\ t=\text{time}\\ a=\text{acceleration}\end{array}$

Calculation:

Consider the motion of the baseball in vertical direction.

Substituting the values, to find the time.

  $\begin{array}{l}0=22t+(0.5)(-9.8)t^2\\ t=4.5s\end{array}$

Conclusion:

Therefore, time spent by baseball in air is 4.5 seconds.