Chapter 2, Problem 83P

(a)

To determine

At what time does the ball reach its maximum height?

Answer to Problem 83P

The ball reach its maximum height at $0.30\text{s}$.

Explanation of Solution

The graph shown plots the velocity-time graph of the ball. When the ball reaches the maximum height, its velocity will be zero there. From the graph at time $0.30\text{s}$ the velocity is zero. Therefore it is the time when the ball reaches the maximum for the first time.

Conclusion

Therefore, The ball reach its maximum height at $0.30\text{s}$.

(b)

To determine

For how long is the ball in contact with the floor?

Answer to Problem 83P

The ball remains in contact with the floor for time of $0.05\text{s}$.

Explanation of Solution

The graph shown plots the velocity-time graph of the ball. The time for which the ball remains in contact with the floor is equal to the time for which the ball makes a transition from its extreme negative velocity point to extreme positive velocity.

Write the equation to find the time of transition of ball.

$t=t_2-t_1$                 (I)

Here, $t$ is the time of transition of ball from negative velocity to positive velocity, $t_2$ is the time for which ball reaches positive velocity, and $t_1$ is the time at which ball reaches negative velocity.

Conclusion:

Substitute $0.65\text{s}$ for $t_2$ and $0.60\text{s}$ for $t_1$ in equation (I) to get $t$.

$\begin{array}{c}t=0.65\text{s}-0.60\text{s}\\ \text{=0}\text{.05}\text{s}\end{array}$

Therefore, The ball remains in contact with the floor for time of $0.05\text{s}$.

(c)

To determine

What is the maximum height reached by the ball?

Answer to Problem 83P

The maximum height reached by ball is $\text{0}\text{.45}\text{m}$.

Explanation of Solution

Write the equation to find the maximum height reached by the rocket.

$y_f=v_{iy}\Delta t\frac{1}{2}{a}_y{\left(\Delta t\right)}^2$ (I)

Here, $y_f$ is the maximum height, $v_{iy}$ is the initial velocity, $\Delta t$ is the change in time, $a_y$ is acceleration

Substitute $\frac{\Delta v_y}{\Delta t}$ for $a_y$  in equation (I).

$y_f=v_{iy}\left(\Delta t\right)+\frac{1}{2}\frac{\Delta v_y}{\Delta t}{\left(\Delta t\right)}^2$ (II)

Conclusion:

Substitute $3\text{m}/\text{s}$ for $v_{iy}$ , $0.3\text{s}$ for $\Delta t$ , $3\text{m}/\text{s}$ for $\Delta v_y$ in equation (II) to get $y_f$.

$\begin{array}{c}{y}_f=\left(3.0\text{m}/\text{s}\right)\left(0.30\text{s}\right)+\frac{0-3\text{m}/\text{s}}{2}\left(0.30\text{s}\right)\\ =0.45\text{m}\end{array}$

Therefore, The maximum height reached by ball is $\text{0}\text{.45}\text{m}$.

(d)

To determine

What is the acceleration of the ball while in the air?

Answer to Problem 83P

The acceleration of the ball while in the air is $10\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Write the equation to find the acceleration of the ball.

$\begin{array}{c}{a}_y=\frac{\Delta v_y}{\Delta t}\\ =\frac{v_2-v_1}{\Delta t}\end{array}$ (I)

Here, $a_y$ is the acceleration of the ball, is the final velocity , $v_1$ is the initial velocity and $\Delta t$ is the change in time

Conclusion:

Substitute $-3\text{m}/\text{s}$ for $v_2$ , $3\text{m}/\text{s}$ for $v_1$ , $0.60\text{s}$ for $\Delta t$ in equation (I) to get $a_y$.

$\begin{array}{c}{a}_y=\frac{-3\text{m}/\text{s}-3\text{m}/\text{s}}{0.60\text{s}}\\ =-10\text{m}/{\text{s}}^{\text{2}}\end{array}$

Since the acceleration is negative, it is acting down wards.

Therefore, The acceleration of the ball while in the air is $10\text{m}/{\text{s}}^{\text{2}}$.

(e)

To determine

What is the average acceleration of the floor while in contact with the floor?

Answer to Problem 83P

The average acceleration of the ball is $120\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Write the equation to find the average acceleration of the ball.

$\begin{array}{c}{a}_{av}=\frac{\Delta v}{\Delta t}\\ =\frac{v_2-v_1}{\Delta t}\end{array}$ (I)

Here, $a_{av}$ is the average acceleration of the ball, is the final velocity , $v_1$ is the initial velocity and $\Delta t$ is the change in time

Conclusion:

Substitute $3\text{m}/\text{s}$ for $v_2$ , $-3\text{m}/\text{s}$ for $v_1$ , $0.05\text{s}$ for $\Delta t$ in equation (I) to get $a_y$.

$\begin{array}{c}{a}_y=\frac{3\text{m}/\text{s}-\left(-3\text{m}/\text{s}\right)}{0.05\text{s}}\\ =120\text{m}/{\text{s}}^{\text{2}}\end{array}$

Since the acceleration is positive, it is acting upwards.

Therefore, The average acceleration of the ball is $120\text{m}/{\text{s}}^{\text{2}}$.