Chapter 2, Problem 48P

(a)

To determine

The initial speed of baseball.

Answer to Problem 48P

The initial speed of baseball is $26.8\text{ m/s}$ .

Explanation of Solution

Given:

Final velocity, $v=13\text{ m/s}$

Height, $h=28\text{m}$

Formula used:

  $v^2=u^2-2gh$

Where, $h$ = height

  $g$ = acceleration due to gravity

  $u$ = initial velocity

  $v$ = final velocity

Calculation:

The initial velocity is calculated as,

  $\begin{array}{l}{v}^2=u^2-2gh\\ {\left(13\right)}^2=u^2-2\left(9.8\right)\left(28\right)\\ u^2={\left(13\right)}^2+2\left(9.8\right)\left(28\right)\\ u=\sqrt{{\left(13\right)}^2+2\left(9.8\right)\left(28\right)}\\ u=26.8\text{ m/s}\end{array}$

Conclusion:

The initial speed of the ball is $26.8\text{ m/s}$ .

(b)

To determine

The maximum height attained by the ball.

Answer to Problem 48P

The maximum height attained by the ball is $36.64\text{ m}$ .

Explanation of Solution

Given:

Initial speed, $u=26.8\text{ m/s}$

Final speed, $v=0\text{ m/s}$

Formula used:

  $H=\frac{u^2}{2g}$

Where, $H$ = maximum height

Calculation:

The maximum height attained by the ball is,

  $\begin{array}{l}H=\frac{u^2}{2g}\\ H=\frac{{\left(26.8\right)}^2}{2\left(9.8\right)}\\ H=36.64\text{ m}\end{array}$

Conclusion:

  $H=36.64\text{ m}$

(c)

To determine

Time taken by ball to reach the window from initial point.

Answer to Problem 48P

The time taken by ball to reach at the window is $1.4\text{ s}$ from ground.

Explanation of Solution

Given:

Initial speed, $u=26.8\text{ m/s}$

Final speed, $v=13\text{ m/s}$

Formula used:

  $v=u-gt$

Calculation:

The time taken by ball to reach at window is calculated as,

  $\begin{array}{l}\text{13}=26.8-9.8t\\ t=\frac{26.8-13}{9.8}\\ t=1.4\text{ s}\end{array}$

Conclusion:

  $t=1.4\text{ s}$

(d)

To determine

Time taken by ball to reach at the initial point from window.

Answer to Problem 48P

In $\text{4}\text{.09 s}$ the ball will reach at its initial point from window.

Explanation of Solution

Given:

Initial velocity, $u=26.8\text{ m/s}$

Displacement, $h=0$

Formula used:

  $h=ut-\frac{g{t}^2}{2}$

Calculation:

The taken by ball to reach ground after it thrown from ground (or street) is calculated as,

  $\begin{array}{l}h=ut-\frac{g{t}^2}{2}\\ 0=\left(26.8\right)t-\frac{9.8t^2}{2}\\ t=\frac{26.8}{4.9}\\ t=5.5\text{ s}\end{array}$

Time taken by ball to reach the ground after it passed the window is,

  $\begin{array}{l}T=T_{total}-T_{window}\\ T=5.5-1.4\\ T=4.09\text{ s}\end{array}$

Conclusion:

  $T=4.09\text{ s}$