Chapter 7, Problem 7.1DQ

A baseball is thrown straight up with initial speed ν₀. If air resistance cannot be ignored, when the ball returns to its initial height its speed is less than ν₀. Explain why, using energy concepts.

To determine

The explanation using energy concept for when the ball return to its initial height, its speed is less than $v_0$ if air resistance is not ignored.

Explanation of Solution

Section 1:

To determine: The velocity of the ball when it returns to its initial height if the air resistance is not ignored.

Introduction: Air resistance opposes the motion of the baseball while it is moving downward.

Formula to calculate the initial potential energy of the ball is,

$U_{\text{i}}=mg{h}_{\text{i}}$

• $U_{\text{i}}$ is the initial potential energy of the ball.
• $m$ is the mass of the ball.
• $g$ is the gravitational acceleration.
• $h_{\text{i}}$ is the initial height of the ball.

Formula to calculate the initial kinetic energy of the ball is,

$K_{\text{i}}=\frac{1}{2}m{v}_{\text{i}}{}^2$

• $K_{\text{i}}$ is the initial kinetic energy of the ball.
• $v_{\text{i}}$ is the initial velocity of the ball.

Formula to calculate the final potential energy of the ball is,

$U_{\text{f}}=mg{h}_{\text{f}}$

• $U_{\text{f}}$ is the final potential energy of the ball.
• $h_{\text{f}}$ is the final height of the ball.

Formula to calculate the final kinetic energy of the ball is,

$K_{\text{f}}=\frac{1}{2}m{v}_{\text{f}}{}^2$

• $K_{\text{f}}$ is the final kinetic energy of the ball.
• $v_{\text{f}}$ is the final velocity of the ball.

Expression for law of conservation of energy is,

$U_{\text{i}}+K_{\text{i}}=U_{\text{f}}+K_{\text{f}}$

Substitute $mg{h}_{\text{i}}$ for $U_{\text{i}}$ , $\frac{1}{2}m{v}_{\text{i}}{}^2$ for $K_{\text{i}}$ , $mg{h}_{\text{f}}$ for $U_{\text{f}}$ and $\frac{1}{2}m{v}_{\text{f}}{}^2$ for $K_{\text{f}}$ to find $v_{\text{f}}$.

$mg{h}_{\text{i}}+\frac{1}{2}m{v}_{\text{i}}{}^2=mg{h}_{\text{f}}+\frac{1}{2}m{v}_{\text{f}}{}^2$

Substitute $h$ for $h_{\text{i}}$, $0\text{m}/\text{s}$ for $v_{\text{i}}$, $0\text{m}$ for $h_{\text{f}}$ and $v_0$ for $v_{\text{f}}$ to find $v_0$.

$\begin{array}{c}mg\left(h\right)+\frac{1}{2}m\times {\left(0\text{m}/\text{s}\right)}^2=mg\left(0\text{m}\right)+\frac{1}{2}m{\left(v_0\right)}^2\\ mgh+0=0+\frac{1}{2}m{v}_0{}^2\\ gh=\frac{1}{2}{v}_0{}^2\\ v_0=\sqrt{2gh}\end{array}$ (I)

Section 2:

To determine: The velocity of the ball when it returns to its initial height if the air resistance cannot be ignored.

Introduction: Air resistance opposes the motion of the baseball while it is moving downward.

If air resistance cannot be ignored, some amount of work has to be done by the ball against the air friction force. Therefore,

In this case, expression for law of conservation of energy is,

$U_{\text{i}}+K_{\text{i}}+{\left(-W\right)}_{\text{f}}=U_{\text{f}}+K_{\text{f}}$

• $U_{\text{i}}$ is the initial potential energy of the ball.
• $K_{\text{i}}$ is the initial kinetic energy of the ball.
• $U_{\text{f}}$ is the final potential energy of the ball.
• $K_{\text{f}}$ is the final kinetic energy of the ball.
• $W_{\text{f}}$ is the work done against the air friction force.

Substitute $mg{h}_{\text{i}}$ for $U_{\text{i}}$ , $\frac{1}{2}m{v}_{\text{i}}{}^2$ for $K_{\text{i}}$ , $mg{h}_{\text{f}}$ for $U_{\text{f}}$ and $\frac{1}{2}m{v}_{\text{f}}{}^2$ for $K_{\text{f}}$ to find $v_{\text{f}}$.

$mg{h}_{\text{i}}+\frac{1}{2}m{v}_{\text{i}}{}^2+\left(-W_{\text{f}}\right)=mg{h}_{\text{f}}+\frac{1}{2}m{v}_{\text{f}}{}^2$

Substitute $h$ for $h_{\text{i}}$ , $0\text{m}/\text{s}$ for $v_{\text{i}}$ , $0\text{m}$ for $h_{\text{f}}$ and $v_0\text{'}$ for $v_{\text{f}}$ to find $v_0\text{'}$.

$\begin{array}{c}mg\left(h\right)+\frac{1}{2}m\times {\left(0\text{m/s}\right)}^2+\left(-W_{\text{f}}\right)=mg\times \left(0\text{m}\right)+\frac{1}{2}m{\left(v_0\text{'}\right)}^2\\ mgh+0+\left(-W_{\text{f}}\right)=0+\frac{1}{2}m{v}_0{\text{'}}^2\\ gh+\left(-\frac{W_{\text{f}}}{m}\right)=\frac{1}{2}{v}_0{\text{'}}^2\\ v_0\text{'}=\sqrt{2\left(gh-\frac{W_{\text{f}}}{m}\right)}\end{array}$ (II)

From Equation (I) and Equation (II),

$v_0\text{'}<v_0$

Conclusion:

Therefore, when the ball returns to its initial height its speed is less than $v_0$ as some amount of has been done by the ball against the friction which causes the reduction in its speed.