Question

A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he releases it. Use conversation of energy to find

Part A

The ball's maximum height above the ground.

Express your answer using two significant figures.

hmax =		m

Part B

The ball's speed as it passes the window on its way down.

Express your answer using two significant figures.

vwindow =		m/s

Part C

The speed of impact on the ground.

Express your answer using two significant figures.

vimpact =		m/s

Answer 1

Concepts and reason

The concept required to solve this problem is conservation of energy.

Initially, use the conservation of energy expression to calculate the maximum height to which the ball goes.

Later use the conservation of energy expression to calculate the speed of the ball as it passes the window on its way down.

Finally, use the conservation of energy expression to calculate the speed of the impact of the ball to the ground.

Fundamentals

The conservation of energy theorem states that the sum of initial potential energy and kinetic energy to the sum of final potential energy and kinetic energy. The equation of conservation of energy is,

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

Here, $U$ is the gravitational potential energy, $K$ is the kinetic energy, and subscripts ${\rm{ i and f}}$ represent initial and final state.

The gravitational potential energy is,

$U = mgh$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $h$ is the height of the ball.

The kinetic energy is,

$K = \frac{1}{2}m{v^2}$

Here, $m$ is mass, and $v$ is the speed of the ball.

Part A

Use the conservation of energy equation.

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

Substitute $mg{h_{\rm{i}}}$ for ${U_{\rm{i}}}$ , $\frac{1}{2}m{v^2}$ for ${K_{\rm{i}}}$ , $mg{h_{\max }}$ for ${U_{\rm{f}}}$ , and $0$ for ${K_{\rm{f}}}$ in the equation ${U_{\rm{i}}} + {K_{\rm{i}}} = {U_{\rm{f}}} + {K_{\rm{f}}}$ and solve for maximum height ${h_{\max }}$ .

$\begin{array}{c}\\mg{h_{\rm{i}}} + \frac{1}{2}m{v^2} = mg{h_{\max }} + 0\\\\{h_{\max }} = \frac{{2gh + {v^2}}}{{2g}}\\\end{array}$

Substitute $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ , $20{\rm{ m}}$ for ${h_{\rm{i}}}$ , and $10{\rm{ m/s}}$ for $v$ in the equation ${h_{\max }} = \frac{{2gh + {v^2}}}{{2g}}$ and calculate the maximum height ${h_{\max }}$ .

$\begin{array}{c}\\{h_{\max }} = \frac{{2\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {20{\rm{ m}}} \right) + {{\left( {10{\rm{ m/s}}} \right)}^2}}}{{2\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}\\\\ = 25{\rm{ m}}\\\end{array}$

Part B

Use the conservation of energy equation.

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

Substitute $mg{h_{\rm{i}}}$ for ${U_{\rm{i}}}$ , $\frac{1}{2}m{v^2}$ for ${K_{\rm{i}}}$ , and $mg{h_{\rm{i}}}$ for ${U_{\rm{f}}}$ in the equation ${U_{\rm{i}}} + {K_{\rm{i}}} = {U_{\rm{f}}} + {K_{\rm{f}}}$ and solve for kinetic energy as it passes the window on its way down.

$\begin{array}{c}\\mg{h_{\rm{i}}} + \frac{1}{2}m{v^2} = mg{h_{\rm{i}}} + {K_{\rm{f}}}\\\\{K_{\rm{f}}} = \frac{1}{2}m{v^2}\\\end{array}$

As the initial kinetic energy and final kinetic energies are same as ball passes the window on its way down so the speed of the ball is same as initial speed that is ${v_{{\rm{window}}}} = 10{\rm{ m/s}}$ .

Part C

Use the conservation of energy equation.

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

Substitute $mg{h_{{\rm{max}}}}$ for ${U_{\rm{i}}}$ , $0$ for ${K_{\rm{i}}}$ , $0$ for ${U_{\rm{f}}}$ , and $\frac{1}{2}mv_{{\rm{impact}}}^2$ for ${K_{\rm{f}}}$ in the equation ${U_{\rm{i}}} + {K_{\rm{i}}} = {U_{\rm{f}}} + {K_{\rm{f}}}$ and solve for speed of impact ${v_{{\rm{impact}}}}$ .

$\begin{array}{c}\\mg{h_{\max }} + 0 = 0 + \frac{1}{2}mv_{{\rm{impact}}}^2\\\\{v_{{\rm{impact}}}} = \sqrt {2gh} \\\end{array}$

Substitute $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ , and $25{\rm{ m}}$ for ${h_{{\rm{max}}}}$ in the equation ${v_{{\rm{impact}}}} = \sqrt {2gh}$ and calculate the speed of impact ${v_{{\rm{impact}}}}$ .

$\begin{array}{c}\\{v_{{\rm{impact}}}} = \sqrt {2\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {25{\rm{ m}}} \right)} \\\\ = 22{\rm{ m/s}}\\\end{array}$

Ans: Part A

The ball's maximum height above the ground is $25{\rm{ m}}$ .