Chapter 2, Problem 66AP

Two students air on a balcony a distance h above the street. One student throws a ball vertically down-ward at a speed v₀; at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of v₀, g, h, and t. (a) Write the kinematic equation for the y-coordinate of each ball, (b) Set the equations found in part (a) equal to height 0 and solve each for t symbolically using the quadratic formula. What is the difference in the two balls' time in the air? (c) Use the time-independent kinematics equation to find the velocity of each ball as it strikes the ground, (d) How far apart are the balls at a time t after they are released and before the strike the ground?

To determine

The kinematic equation of ball thrown downwards.

Answer to Problem 66AP

Solution: The equation is $y_1=h-v_{0}t-\frac{1}{2}g{t}^2$.

Explanation of Solution

Given Info: Initial velocity of ball is $v_0$.

The displacement and velocity in downward direction is negative and in upward direction is taken as positive. The acceleration due to gravity is acting vertically downwards in both upward and downward motion.

Write the formula to calculate the displacement of ball thrown vertically downwards from the balcony.

$y_0-h=-v_{0}t+\frac{1}{2}\left(-g\right)t^2$

• $y_1$ is the displacement of the ball thrown vertically downwards
• $h$ is the distance of balcony from the street
• $v_0$ is the initial velocity of ball
• $g$ is the acceleration due to gravity
• $t$ is the time duration for displacement $y_0$

Rearrange the above equation.

$y_1=h-v_{0}t-\frac{1}{2}g{t}^2$ (I)

The displacement and velocity in downward direction is negative and in upward direction is taken as positive. The acceleration due to gravity is acting vertically downwards in both upward and downward motion.

Write the formula to calculate the displacement of ball thrown vertically upwards from the balcony.

$y_2-h=v_{0}t+\frac{1}{2}\left(-g\right)t^2$

• $y_2$ is the displacement of the ball thrown vertically upwards

Rearrange the above equation.

$y_2=h+v_{0}t-\frac{1}{2}g{t}^2$ (II)

Thus, the kinematic equation for ball thrown vertically upwards will be

$y_2=h+v_{0}t-\frac{1}{2}g{t}^2$.

Conclusion:

Thus, equation for ball thrown in vertically downward and upward directions are $y_1=h-v_{0}t-\frac{1}{2}g{t}^2$ and $y_2=h+v_{0}t-\frac{1}{2}g{t}^2$ respectively.

To determine

Solve the quadratic equations by setting height equals zero and calculate the time difference of two balls in air.

Answer to Problem 66AP

Solution: the solutions of quadratic equations and the time difference between the balls in air are, $t_1=-\frac{v_0}{g}+\sqrt{{\left(\frac{v_0}{g}\right)}^2+\frac{2h}{g}}$, $t_2=\frac{v_0}{g}+\sqrt{{\left(\frac{v_0}{g}\right)}^2+\frac{2h}{g}}$.and $\frac{2v_0}{g}$ respectively

Explanation of Solution

Given Info: Initial velocity of ball is $v_0$.

Let $t_1$ be the duration of the ball 1 in air.

Write the formula to calculate the displacement of ball thrown vertically downwards from the balcony.

$y_1=h-v_0{t}_1-\frac{1}{2}g{t}_1{}^2$

Substitute $0$ for $y_1$ in the above equation to find $t_1$.

$0=h-v_0{t}_1-\frac{1}{2}g{t}_1{}^2$

Multiply each term by $2$ and rearrange the above equation in the standard form of a quadratic equation.

$g{t}_1{}^2+2v_0{t}_1+\left(-2h\right)=0$

The solution of above equation can be found using quadratic formula.

$\begin{array}{c}{t}_1=\frac{-2v_0\pm \sqrt{{\left(2v_0\right)}^2-4g\left(-2h\right)}}{2g}\\ =-\frac{v_0}{g}\pm \sqrt{{\left(\frac{v_0}{g}\right)}^2+\frac{2h}{g}}\end{array}$

Consider only the positive solution of $t_1$.

$t_1=-\frac{v_0}{g}+\sqrt{{\left(\frac{v_0}{g}\right)}^2+\frac{2h}{g}}$ (III)

Let $t_2$ be the duration of the ball 2 in air.

Write the formula to calculate the displacement of ball thrown vertically upwards from the balcony.

$y_2=h+v_0{t}_1-\frac{1}{2}g{t}_1{}^2$

Substitute $0$ for $y_2$ $y_1$ in the above equation to find $t_2$.

$0=h+v_0{t}_1-\frac{1}{2}g{t}_1{}^2$

Multiply each term by $2$ and rearrange the above equation in the standard form of a quadratic equation.

$g{t}_2{}^2+\left(-2v_0{t}_2\right)+\left(-2h\right)=0$

The solution of above equation can be found using quadratic formula.

$\begin{array}{c}{t}_2=\frac{-\left(-2v_0\right)\pm \sqrt{{\left(-2v_0\right)}^2-4g\left(-2h\right)}}{2g}\\ =\frac{v_0}{g}\pm \sqrt{{\left(\frac{v_0}{g}\right)}^2+\frac{2h}{g}}\end{array}$

Consider only the positive solution of $t_2$.

$t_2=\frac{v_0}{g}+\sqrt{{\left(\frac{v_0}{g}\right)}^2+\frac{2h}{g}}$ (IV)

Thus, the value of $t_2$ is $\frac{v_0}{g}+\sqrt{{\left(\frac{v_0}{g}\right)}^2+\frac{2h}{g}}$.

Conclusion:

Write the time difference of two balls in air.

$\Delta t=t_2-t_1$

• $\Delta t$ is the time difference between two balls in air

Substitute equation (IV) for $t_2$ and equation (III) for $t_1$ in the above equation to find $\Delta t$.

$\begin{array}{c}\Delta t=\frac{v_0}{g}+\sqrt{{\left(\frac{v_0}{g}\right)}^2+\frac{2h}{g}}-\left(-\frac{v_0}{g}+\sqrt{{\left(\frac{v_0}{g}\right)}^2+\frac{2h}{g}}\right)\\ =\frac{v_0}{g}+\frac{v_0}{g}+\sqrt{{\left(\frac{v_0}{g}\right)}^2+\frac{2h}{g}}-\sqrt{{\left(\frac{v_0}{g}\right)}^2+\frac{2h}{g}}\\ =\frac{2v_0}{g}\end{array}$

Thus, the time difference between the balls in air $\frac{2v_0}{g}$ respectively.

To determine

The final velocity of balls on touching the ground.

Answer to Problem 66AP

Solution: The velocity of two balls on touching the ground will be $-\sqrt{v_0{}^2+2gh}$.

Explanation of Solution

Given Info: Initial velocity of ball is $v_0$.

The displacement and velocity in downward direction is negative and in upward direction is taken as positive. The acceleration due to gravity is acting vertically downwards in both upward and downward motion.

Write the formula to calculate the final velocity.

$v_{\text{f1}}=-\sqrt{v_0{}^2+2\left(-g\right)\left(-h\right)}$

• $v_{\text{f1}}$ is the final velocity of ball thrown vertically downwards

Rewrite the above equation as follows.

$v_{\text{f1}}=-\sqrt{v_0{}^2+2gh}$

The displacement and velocity in downward direction is negative and in upward direction is taken as positive. The acceleration due to gravity is acting vertically downwards in both upward and downward motion.

Write the formula to calculate the velocity of ball thrown upwards on touching the ground.

$v_{\text{f2}}=-\sqrt{v_0{}^2+2\left(-g\right)\left(-h\right)}$

• $v_{\text{f2}}$ is the velocity of ball thrown vertically upwards at the moment of touching the ground.

Rewrite the above equation as follows.

$v_{\text{f2}}=-\sqrt{v_0{}^2+2gh}$

Conclusion: Thus, the final velocities of two balls are same and the value is $-\sqrt{v_0{}^2+2gh}$.

To determine

The separation between balls at time $t$.

Answer to Problem 66AP

Solution: The separation between balls at time t will be $2v_{0}t$.

Explanation of Solution

Given Info: Initial velocity of ball is $v_0$.

Write the formula to calculate the separation between the balls.

$\Delta y=y_2-y_1$

• $\Delta y$ is the separation between balls at time $t$

Substitute equation (II) for $y_2$ and equation (I) for $y_1$ in the above equation.

$\begin{array}{c}\Delta y=\left(h+v_0{t}_2-\frac{1}{2}g{t}_2{}^2\right)-\left(h-v_0{t}_1-\frac{1}{2}g{t}_1{}^2\right)\\ =v_0\left(t_2+t_1\right)\end{array}$

Substitute $t$ for $t_1$ and $t_2$ in the above equation to find $\Delta y$.

$\begin{array}{c}\Delta y=v_0\left(t+t\right)\\ =2v_{0}t\end{array}$

Conclusion:

Thus, the separation between balls at time t will be $2v_{0}t$