Chapter 3, Problem 61P

(a)

To determine

The maximum height above the ground reached by the cannonball.

Answer to Problem 61P

The maximum height above the ground reached by the cannonball is $37\text{ m}$ .

Explanation of Solution

Take the $+y$ direction to be upward.

Write the equation of motion.

$v_{fy}^2-v_{iy}^2=2a_y\Delta y$ (I)

Here, $v_{fy}$ is the $y$ component of final velocity, $v_{iy}$ is the $y$ component of initial velocity, $a_y$ is the acceleration in $y$ direction and $\Delta y$ is the displacement in $y$ direction

When the cannon ball reaches maximum height, its velocity will be zero.

$v_{fy}=0$ (II)

Write the equation for $v_{iy}$ .

$v_{iy}=v_i\sin \theta$ (III)

Here, $v_i$ is the initial velocity of cannonball and $\theta$ is the angle the cannonball makes with the horizontal initially

The acceleration of the ball will be the negative of acceleration due to gravity.

$a_y=-g$ (IV)

Write the equation for $\Delta y$ .

$\Delta y=y_f-y_i$ (V)

Here, $y_f$ is the maximum height reached by the cannonball and $y_i$ is the initial height of the cannonball

Put equations (II) to (V) in equation (I) and rewrite it for $y_f$ .

$\begin{array}{c}0-{\left(v_i\sin \theta \right)}^2=2\left(-g\right)\left(y_f-y_i\right)\\ v_i^2{\sin }^2\theta =2g\left(y_f-y_i\right)\\ 2g{y}_f=2g{y}_i+v_i^2{\sin }^2\theta \\ y_f=y_i+\frac{v_i^2{\sin }^2\theta }{2g}\end{array}$ (VI)

Conclusion:

Given that the initial height of the cannonball is $7.0\text{ m}$ , its initial velocity is $40\text{ m/s}$ and the angle the cannonball makes initially with the horizontal is $37°$ .

The value of acceleration due to gravity is $9.80{\text{ m/s}}^2$ .

Substitute $7.0\text{ m}$ for $y_i$ , $40\text{ m/s}$ for $v_i$ , $37°$ for $\theta$ and $9.80{\text{ m/s}}^2$ for $g$ in equation (VI) to find $y_f$ .

$\begin{array}{c}{y}_f=7.0\text{ m}+\frac{\left(40\text{ m/s}\right){\sin }^{2}37°}{2\left(9.80{\text{ m/s}}^2\right)}\\ =7.0\text{ m}+30\text{ m}\\ =37\text{ m}\end{array}$

Therefore, the maximum height above the ground reached by the cannonball is $37\text{ m}$ .

(b)

To determine

The horizontal distance from the release point at which the cannonball will land if it makes it over the castle walls and lands back down on the ground.

Answer to Problem 61P

The horizontal distance from the release point at which the cannonball will land if it makes it over the castle walls and lands back down on the ground is $170\text{ m}$ .

Explanation of Solution

Write the equation of motion in $y$ direction.

$\Delta y=v_{iy}\Delta t+\frac{1}{2}{a}_y{\left(\Delta t\right)}^2$ (VII)

Here, $\Delta t$ is the time taken for the motion

Write the equation for the horizontal distance at which the cannonball lands.

$\Delta x=v_x\Delta t$

Here, $\Delta x$ is the horizontal distance at which the cannonball lands and $v_x$ is the velocity in $x$ direction

Rewrite the above equation for $\Delta t$ .

$\Delta t=\frac{\Delta x}{v_x}$ (VIII)

Refer to figure 1 and write the equation for $v_x$ .

$v_x=v_i\cos \theta$ (IX)

Put equation (IX) in equation (VIII).

$\Delta t=\frac{\Delta x}{v_i\cos \theta }$ (X)

Put equations (III), (IV) and (X) in equation (VII).

$\begin{array}{c}\Delta y=v_i\sin \theta \frac{\Delta x}{v_i\cos \theta }-\frac{1}{2}g\frac{{\left(\Delta x\right)}^2}{v_i^2{\cos }^2\theta }\\ =\Delta x\tan \theta -\frac{g{\left(\Delta x\right)}^2}{2v_i^2{\cos }^2\theta }\\ \frac{g}{2v_i^2{\cos }^2\theta }{\left(\Delta x\right)}^2-\left(\tan \theta \right)\Delta x+\Delta y=0\end{array}$

Solve the above quadratic formula in $\Delta x$ .

$\begin{array}{c}\Delta x=\frac{\tan \theta \pm \sqrt{{\tan }^2\theta -\frac{4g\Delta y}{2v_i^2{\cos }^2\theta }}}{\frac{2g}{2v_i^2{\cos }^2\theta }}\\ =\frac{\tan \theta \pm \sqrt{{\tan }^2\theta -\frac{2g\Delta y}{v_i^2{\cos }^2\theta }}}{\frac{g}{v_i^2{\cos }^2\theta }}\end{array}$ (XI)

Conclusion:

When the cannonball reaches the ground the value of $\Delta y$ is $-7.0\text{ m}$ .

Substitute $37°$ for $\theta$, $9.80{\text{ m/s}}^2$ for $g$, $-7.0\text{ m}$ for $\Delta y$ and $40\text{ m/s}$ for $v_i$ , in equation (XI) to find $\Delta x$ .

$\begin{array}{c}\Delta x=\frac{\tan 37°\pm \sqrt{{\tan }^{2}37°-\frac{2\left(9.80{\text{ m/s}}^2\right)\left(-7.0\text{ m}\right)}{{\left(40\text{ m/s}\right)}^2{\cos }^{2}37°}}}{\frac{\left(9.80{\text{ m/s}}^2\right)}{{\left(40\text{ m/s}\right)}^2{\cos }^{2}37°}}\\ =170\text{ m or }-9\text{ m}\end{array}$

The catapult does not fire backward. This means $-9\text{ m}$ is unphysical.

Therefore, the horizontal distance from the release point at which the cannonball will land if it makes it over the castle walls and lands back down on the ground is $170\text{ m}$ .

(c)

To determine

The $x$ and $y$ components of the cannonball’s velocity just before it lands.

Answer to Problem 61P

The $x$ component of the cannonball’s velocity is $32\text{ m/s}$ and the $y$ component is $-27\text{ m/s}$ .

Explanation of Solution

The $x$ component of the cannonball’s velocity is same as the initial value.

$v_{fx}=v_x$

Here, $v_{fx}$ is the $x$ component of the cannonball’s velocity as it lands

Put equation (IX) in the above equation.

$v_{fx}=v_i\cos \theta$ (XII)

Rewrite equation (I) for $v_{fy}^2$ .

$v_{fy}^2=v_{iy}^2+2a_y\Delta y$

Put equations (III) and (IV) in the above equation and rewrite it for $v_{fy}$ .

$\begin{array}{c}{v}_{fy}^2=v_i^2{\sin }^2\theta -2g\Delta y\\ v_{fy}=\sqrt{v_i^2{\sin }^2\theta -2g\Delta y}\end{array}$ (XIII)

Conclusion:

Substitute $40\text{ m/s}$ for $v_i$ and $37°$ for $\theta$ in equation (XII) to find $v_{fx}$ .

$\begin{array}{c}{v}_{fx}=\left(40\text{ m/s}\right)\cos 37°\\ =32\text{ m/s}\end{array}$

Substitute $40\text{ m/s}$ for $v_i$ , $37°$ for $\theta$, $9.80{\text{ m/s}}^2$ for $g$ and $-7.0\text{ m}$ for $\Delta y$ in equation (XIII) to find $v_{fy}$ .

$\begin{array}{c}{v}_{fy}=\sqrt{{\left(40\text{ m/s}\right)}^2{\sin }^{2}37°-2\left(9.80{\text{ m/s}}^2\right)\left(-7.0\text{ m}\right)}\\ =-27\text{ m/s}\end{array}$

The negative sign indicates the velocity is downward.

Therefore, the $x$ component of the cannonball’s velocity is $32\text{ m/s}$ and the $y$ component is $-27\text{ m/s}$ .