Question

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

a.The ball's maximum height above the ground.

b.The ball's speed as it passes the window on its waydown.

c.The speed of impact on the ground.

Answer 1

Concepts and reason

The concept of conservation of energy is used here.

In the first part, equate the initial kinetic potential energy equals to the final kinetic and potential energy at the maximum height. Find the expression of the maximum height above the ground, substitute the values in the expression and find the result.

In the next part, find the ball speed as it passes the window. In this part, height remains the same.

In the last part, at the ground the height is zero. Using the energy conservation, find the speed of impact on the ground.

Fundamentals

The law of conservation of energy states that sum of initial kinetic and potential energy is equal to the sum of final kinetic and potential. The total energy always remains conserved.

Mathematically, it is given as:

$K{E_{\rm{i}}} + P{E_{\rm{i}}} = K{E_{\rm{f}}} + P{E_{\rm{f}}}$

Here, $K{E_{\rm{i}}}$ and $K{E_{\rm{f}}}$ are the initial and final kinetic energies respectively and, $P{E_{\rm{i}}}$ and $P{E_{\rm{f}}}$ represents the initial and final potential energies respectively.

The expression of the kinetic energy of an object having mass m moving with velocity v is given as follows:

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

The expression of the potential energy at height h is given as follows:

$PE = mgh$

(a)

The expression of the initial potential energy when the ball is tossed upwards is given as:

$P{E_{\rm{i}}} = mg{h_0}$

Here, m is the mass of ball, g is the acceleration due to gravity and ${h_0}$ is the initial height.

The initial kinetic energy of the ball is given as,

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

The final potential energy of the ball at the maximum height is given as:

$P{E_{\rm{f}}} = mg{h_{\max }}$

The final kinetic energy is at the maximum height is:

$K{E_{\rm{f}}} = 0$

Use the conservation of energy principle and substitute all the above equation in the expression $K{E_{\rm{i}}} + P{E_{\rm{i}}} = K{E_{\rm{f}}} + P{E_{\rm{f}}}$ .

$\begin{array}{c}\\\frac{1}{2}m{v^2} + mg{h_0} = 0 + mg{h_{\max }}\\\\{h_{\max }} = {h_0} + \frac{{{v^2}}}{{2g}}\\\end{array}$

Substitute $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, $10{\rm{ m/s}}$ for $v$ and 20.0 m for ${h_0}$ in the above equation.

$\begin{array}{c}\\{h_{\max }} = \left( {20.0{\rm{ m}}} \right) + \frac{{{{\left( {10{\rm{ m/s}}} \right)}^2}}}{{2\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}\\\\ = 25.1{\rm{ m}}\\\end{array}$

(b)

The initial potential energy of the ball is given as:

$P{E_{\rm{i}}} = mg{h_0}$

The initial kinetic energy of the ball is given as:

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

The final potential energy of the ball when the ball passes the window is given as:

$P{E_{\rm{f}}} = mgh$

The final kinetic energy as the ball passes the window is:

$K{E_{\rm{f}}} = \frac{1}{2}m{v_{\rm{f}}}^2$

Use the conservation of energy principle and substitute all the above equation in the expression $K{E_{\rm{i}}} + P{E_{\rm{i}}} = K{E_{\rm{f}}} + P{E_{\rm{f}}}$ .

$\frac{1}{2}m{v^2} + mg{h_0} = mgh + \frac{1}{2}m{v_{\rm{f}}}^2$

Since, the ball also has a height of 20 m on its way down it. Therefore,

$h = {h_0}$

Substitute ${h_0}$ for h in the equation $\frac{1}{2}m{v^2} + mg{h_0} = mgh + \frac{1}{2}m{v_{\rm{f}}}^2$ and solve for ${v_{\rm{f}}}$ .

$\begin{array}{c}\\\frac{1}{2}m{v^2} + mg{h_0} = mg{h_0} + \frac{1}{2}m{v_{\rm{f}}}^2\\\\{v^2} = {v_{\rm{f}}}^2\\\\{v_{\rm{f}}} = v\\\end{array}$

Substitute $10{\rm{ m/s}}$ for v in the equation ${v_{\rm{f}}} = v$ .

${v_{\rm{f}}} = 10{\rm{ m/s}}$

(c)

The initial potential energy of the ball is given as:

$P{E_{\rm{i}}} = mg{h_0}$

The initial kinetic energy of the ball is given as:

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

The final potential energy of the ball when ball hits the ground is given as:

$P{E_{\rm{f}}} = 0$

The final kinetic energy when ball hits the ground is:

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

Use the conservation of energy principle and substitute all the above equation in the expression $K{E_{\rm{i}}} + P{E_{\rm{i}}} = K{E_{\rm{f}}} + P{E_{\rm{f}}}$ and solve for v.

$\begin{array}{c}\\mg{h_0} + \frac{1}{2}m{v_0}^2 = 0 + \frac{1}{2}m{v^2}\\\\v = \sqrt {{v_0}^2 + 2g{h_0}} \\\end{array}$

Substitute $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, 10 m/s for ${v_0}$ , and 20 m for ${h_0}$ in the equation $v = \sqrt {{v_0}^2 + 2g{h_0}}$ .

$\begin{array}{c}\\v = \sqrt {{{\left( {10{\rm{ m/s}}} \right)}^2} + 2\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {20.0{\rm{ m}}} \right)} \\\\ = 22.2{\rm{ m/s}}\\\end{array}$

Ans: Part a

The value of maximum height above the ground is 25.1 m.