Chapter 2, Problem 74AP

(a)

To determine

The time interval between the flights of the two balls.

Answer to Problem 74AP

The time interval between the flights of the two balls is $\underset{\_}{2v_i/g}$.

Explanation of Solution

Write the equation for time interval.

  $\left(\Delta t\right)=t_2-t_1$                                                                                                (I)

Here, $\left(\Delta t\right)$ is the time interval, $t_1$ is the time taken by first ball and $t_2$ is the time taken by the ball 2.

Write the expression from kinematics for the first ball.

$y_1-y_{01}=v_{01}{t}_1+\frac{1}{2}a{t}_1^2$ (II)

Here, $y_1$ is the final height for ball 1, $y_{01}$ is the initial height of ball 1, $v_{01}$ is the initial velocity of ball 1, $a$ is the acceleration.

Write the expression from kinematics for the second ball.

$y_2-y_{02}=v_{02}{t}_2+\frac{1}{2}a{t}_2^2$ (III)

Here, $y_2$ is the final height for ball 2, $y_{02}$ is the initial height of ball 2, $v_{02}$ is the initial velocity of ball 2.

Conclusion:

Substitute $0\text{m}$ for $y_1$, $h$ for $y_{01}$, $-v_i$ for $v_{01}$, $\left(-g\right)$ for $a$ in Equation (II) to find $t_1$. Here, $g$ is the gravitational acceleration.

  $\begin{array}{l}\left(0\text{m}\right)-\left(h\right)=-v_i{t}_1+\frac{1}{2}\left(-g\right)t_1^2\\ g{t}_1^2+2v_i{t}_1-2h=0\\ t_1=-\frac{v_i}{g}+\sqrt{{\left(\frac{v_i}{g}\right)}^2+\left(\frac{2h}{g}\right)}\end{array}$

Consider only positive solution.

Substitute $0\text{m}$ for $y_1$, $h$ for $y_{01}$, $v_i$ for $v_{02}$, $\left(-g\right)$ for $a$ in Equation (III) to find $t_2$.

  $\begin{array}{l}\left(0\text{m}\right)-\left(h\right)=v_i{t}_2+\frac{1}{2}\left(-g\right)t_2^2\\ g{t}_2^2-2v_i{t}_2-2h=0\\ t_2=\frac{v_i}{g}+\sqrt{{\left(\frac{v_i}{g}\right)}^2+\left(\frac{2h}{g}\right)}\end{array}$

Substitute $\left[\frac{v_i}{g}+\sqrt{{\left(\frac{v_i}{g}\right)}^2+\left(\frac{2h}{g}\right)}\right]$ for $t_2$, $\left[-\frac{v_i}{g}+\sqrt{{\left(\frac{v_i}{g}\right)}^2+\left(\frac{2h}{g}\right)}\right]$ for $t_1$, in Equation (I) to find $\left(\Delta t\right)$.

$\begin{array}{c}\left(\Delta t\right)=\left[\frac{v_i}{g}+\sqrt{{\left(\frac{v_i}{g}\right)}^2+\left(\frac{2h}{g}\right)}\right]-\left[-\frac{v_i}{g}+\sqrt{{\left(\frac{v_i}{g}\right)}^2+\left(\frac{2h}{g}\right)}\right]\\ =\frac{2v_i}{g}\end{array}$

Thus, the time interval between the flights of the two balls is $\underset{\_}{2v_i/g}$.

(b)

To determine

The velocity of the balls when they strike the ground.

Answer to Problem 74AP

The velocity of the balls when they strike the ground is $\underset{\_}{-\sqrt{v_i^2+2gh}}$.

Explanation of Solution

Write the equation velocity at which the ball 1 strikes the ground.

  $v_{1,f}=-\sqrt{v_{01}^2+2ah}$ (IV)

Here, $v_{1,f}$ is the final velocity for ball 1, $h$ is the height.

Write the equation velocity at which the ball 2 strikes the ground.

  $v_{2,f}=-\sqrt{v_{02}^2+2ah}$                                                                                                (V)

Here, $v_{2,f}$ is the final velocity for ball 2.

Conclusion:

Substitute $-v_i$ for $v_{01}$, $\left(-g\right)$ for $a$, $\left(-h\right)$ for $h$ in Equation (IV) to find $v_{1,f}$.

  $\begin{array}{c}{v}_{1,f}=-\sqrt{{\left(-v_i\right)}^2+\left\{2\left(-g\right)\left(-h\right)\right\}}\\ =-\sqrt{{\left(v_i\right)}^2+2gh}\end{array}$

Substitute $v_i$ for $v_{02}$, $\left(-g\right)$ for $a$, $\left(-h\right)$ for $h$ in Equation (V) to find $v_{1,f}$.

$\begin{array}{c}{v}_{2,f}=-\sqrt{{\left(v_i\right)}^2+\left\{2\left(-g\right)\left(-h\right)\right\}}\\ =-\sqrt{{\left(v_i\right)}^2+2gh}\end{array}$

Thus, the velocity of the balls when they strike the ground is $\underset{\_}{-\sqrt{v_i^2+2gh}}$.

(c)

To determine

The distance between the balls.

Answer to Problem 74AP

The distance between the balls is $2v_{i}t$.

Explanation of Solution

Write the equation of the distance between the balls.

  $d=y_2-y_1$ (VI)

Here, $d$ is the distance between the balls.

Rewrite the expression for the distance between the balls by using equation (II) and (III).

$d=\left(y_{02}+v_{02}t+\frac{1}{2}a{t}^2\right)-\left(y_{01}+v_{01}t+\frac{1}{2}a{t}^2\right)$ (VII)

Conclusion:

Substitute $h$ for $y_{02}$, $v_i$ for $v_{02}$, $h$ for $y_{01}$, $-v_i$ for $v_{01}$, $\left(-g\right)$ for $a$ in Equation (VII) to find $d$.

  $\begin{array}{c}d=\left(h+v_{i}t-\frac{1}{2}g{t}^2\right)-\left(h-v_{i}t-\frac{1}{2}g{t}^2\right)\\ =2v_{i}t\end{array}$

Thus, the distance between the balls is $\underset{\_}{2v_{i}t}$.