Question

A $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{kg}$ cannonball is fired from a cannon with muzzle speed of $\mathbf{1000}\mathbf{m}/\mathbf{s}$ at an angle of $\mathbf{37}\mathbf{°}$ with the horizontal. A second ball is fired at an angle of $\mathbf{90}\mathbf{°}$. Use the isolated system model to find (a) the maximum height reached by each ball and (b) the total mechanical energy of the ball–Earth system at the maximum height for each ball. Let $\mathbf{y}\mathbf{=}\mathbf{0}$ at the cannon.

Short answer

(a) The maximum height that the first and second ball reach is $18479\text{m}$ and $51020\text{m}$respectively.

(b) The total mechanical energy of each ball at maximum height is $1\times {10}^7\text{J}$.

Step-by-step solution

Identification of given data

The mass of cannonball is $m=20\text{kg}$

The muzzle speed of cannonball is $v=1000\text{m}/\text{s}$

The angle for the first cannonball is $\theta =37°$

The angle for the second cannonball is $\alpha =90°$.

Defining projectile motion

The motion of a projectile, making a certain angle with the horizontal, only under the influence of gravity is known as projectile motion. The projectile moves along a plane, making a parabolic path.

Determination of maximum height reached by both cannonball

(a)

The maximum height for the projectile is given as

$h_1=\frac{v^2{\sin }^2\theta }{2g}$

Here, $g$ is the gravitational acceleration and its value is $9.8{\text{m/s}}^{\text{2}}$.

For the first cannonball, substituting the given values in the above equation.

$$\begin{gathered} h_1=\frac{{\left(1000\text{m/s}\right)}^2{\left(\sin 37°\right)}^2}{2\left(9.8{\text{m/s}}^{\text{2}}\right)} \\ {h}_1=18479\text{m} \end{gathered}$$

For the second cannonball, substituting the given values in the above equation.

$$\begin{gathered} h_2=\frac{{\left(1000\text{m/s}\right)}^2{\left(\sin 90°\right)}^2}{2\left(9.8{\text{m/s}}^{\text{2}}\right)} \\ {h}_2=51020\text{m} \end{gathered}$$

Therefore, the maximum height reached by first and second ball is $18479\text{m}$ and $51020\text{m}$.

Determination of total mechanical energy of Earth-ball system at maximum height

(b)

The total mechanical energy of first ball at maximum height is given as

$E_1=mg{h}_1+\frac{1}{2}m{\left(v\cos \theta \right)}^2$

Substitute the given values in above equation.

$$\begin{gathered} E_1=\left(20\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(18479\text{m}\right)+\frac{1}{2}\left(20\text{kg}\right){\left\{\left(1000\text{m}/\text{s}\right)\left(\cos 37°\right)\right\}}^2 \\ {E}_1=1\times {10}^7\text{J} \end{gathered}$$

The total mechanical energy of second ball at maximum height is given as

$E_2=mg{h}_2+\frac{1}{2}m{\left(v\cos \alpha \right)}^2$

Substitute all the values in above equation.

$$\begin{gathered} E_2=\left(20\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(51020\text{m}\right)+\frac{1}{2}\left(20\text{kg}\right){\left\{\left(1000\text{m}/\text{s}\right)\left(\cos 90°\right)\right\}}^2 \\ {E}_2=1\times {10}^7\text{J} \end{gathered}$$

So, the total mechanical energy of both balls at maximum height is $1\times {10}^7\text{J}$.