Chapter 3, Problem 5TP

To determine

(a)

The graph of the ball's vertical velocity versus time.

Answer to Problem 5TP

The graph of the ball's vertical velocity versus time is shown in figure (1).

Explanation of Solution

Given:

The given table is,

Height (m)	Time (sec)
$0.490$	$0.1$
$0.882$	$0.2$
$1.176$	$0.3$
$1.372$	$0.4$
$1.470$	$0.5$
$1.470$	$0.6$
$1.372$	$0.7$

Formula used:

The velocity of the ball is given as,

  $\begin{array}{c}v=\frac{\Delta h}{\Delta t}\\ =\frac{h_2-h_1}{t_2-t_1}\end{array}$

Calculation:

The velocity of ball is calculated as,

  $\begin{array}{c}v=\frac{h_2-h_1}{t_2-t_1}\\ =\frac{0.882\text{m}-0.49\text{m}}{0.2\text{s}-0.1\text{s}}\\ =3.92\text{m}/\text{s}\end{array}$

The table for other time table is given below.

Height (m)	Time (s)	Velocity (m/s)
$0.490$	$0.1$	$0.490$
$0.882$	$0.2$	$3.92$
$1.176$	$0.3$	$2.94$
$1.372$	$0.4$	$1.96$
$1.470$	$0.5$	$0.98$
$1.470$	$0.6$	$0$
$1.372$	$0.7$	$-0.98$

The graph of the ball's vertical velocity versus time is shown below.

  

  Figure (1)

Conclusion:

The graph of the ball's vertical velocity versus time is shown in figure (1).

To determine

(b)

The graph of the ball's horizontal velocity.

Answer to Problem 5TP

The graph of the ball's horizontal velocity versus time is shown in figure (2).

Explanation of Solution

Formula used:

The acceleration of ball is given as,

  $\begin{array}{c}a=\frac{\Delta v}{\Delta t}\\ =\frac{v_2-v_1}{t_2-t_1}\end{array}$

The vertical velocity of ball is given as,

  $v_{\text{y}}=u_{\text{y}}+at$

Calculation:

The acceleration of ball is calculated as,

  $\begin{array}{c}v=\frac{h_2-h_1}{t_2-t_1}\\ =\frac{3.92\text{m}/\text{s}-4.90\text{m}/\text{s}}{0.2\text{s}-0.1\text{s}}\\ =-9.80\text{m}/{\text{s}}^2\end{array}$

Assume the ball is thrown with initial velocity of $u$ at angle of $60°$ with respect to ground. So the initial horizontal velocity is,

  $\begin{array}{c}{u}_{\text{x}}=u\cos 60°\\ =\frac{u}{2}\end{array}$

The initial vertical velocity is,

  $\begin{array}{c}{u}_{\text{y}}=u\sin 60°\\ =\frac{\sqrt{3}u}{2}\end{array}$

The vertical velocity of ball after $1\text{s}$ is calculated as,

  $\begin{array}{c}4.9\text{m}/\text{s}=\frac{\sqrt{3}u}{2}+\left(-9.80\text{m}/{\text{s}}^2\right)\left(1\text{s}\right)\\ \frac{\sqrt{3}u}{2}=\left(4.9\text{m}/\text{s}+9.80\text{m}/\text{s}\right)\\ u=\frac{2}{\sqrt{3}}\left(14.7\text{m}/\text{s}\right)\\ =16.97\text{m}/\text{s}\end{array}$

The initial horizontal velocity is calculated as,

  $\begin{array}{c}{u}_{\text{x}}=\frac{16.97\text{m}/\text{s}}{2}\\ =8.49\text{m}/\text{s}\end{array}$

The graph of the ball's horizontal velocity versus time is shown below.

  

  Figure (2)

Conclusion:

The graph of the ball's horizontal velocity versus time is shown in figure (2).

To determine

(c)

The graph of ball's vertical acceleration versus time.

Answer to Problem 5TP

The graph of ball's vertical acceleration versus time is shown in figure (3).

Explanation of Solution

Calculation:

The graph of the ball's acceleration versus time is shown below.

  

  Figure (3)

Conclusion:

The graph of ball's vertical acceleration versus time is shown in figure (3).