Chapter 4, Problem 51P

(a)

To determine

The cannonball reaches the maximum height above the ground.

Answer to Problem 51P

The cannonball reaches the maximum height above the ground is $\underset{\_}{36.09\text{m}}$.

Explanation of Solution

Write the expression for the horizontal initial velocity of the cannonball,

  $v_{x0}=v_0\cos \theta$        (I)

Here, $v_{x0}$ is the horizontal initial velocity of the cannonball, $v_0$ is the initial velocity of the cannon ball and $\theta$ is the launch angle.

Write the expression for the vertical initial velocity of the cannonball,

  $v_{y0}=v_0\sin \theta$        (II)

Here, $v_{y0}$ is the vertical initial velocity of the cannonball, $v_0$ is the initial velocity of the cannonball and $\theta$ is the launch angle.

Write the expression for the maximum height of the cannon ball above the ground,

    $h_{\max }=h_0+\left(\frac{v_{yf}^2-v_{y0}^2}{2g}\right)$        (III)

Here, $h_{\max }$ is the maximum height of the cannonball above the ground, $h_0$ is the initial height of the cannonball above the ground, $v_{yf}$ is the final vertical velocity of the cannonball and $g$ is the acceleration due to gravity.

Conclusion:

Substitute $40\text{m/s}$ for $v_0$ and ${37}^{\circ }$ for $\theta$ in (II),

  $\begin{array}{c}{v}_{y0}=\left(40\text{m/s}\right)\sin {37}^{\circ }\\ =24.07\text{m/s}\end{array}$

Substitute $24.07\text{m/s}$ and $v_{y0}$, $0$ for $v_{yf}$ , $7.0\text{m}$ for $h_0$ and $-9.8{\text{m/s}}^{\text{2}}$ for $g$ in (III),

  $\begin{array}{c}{h}_{\max }=7.0\text{m}+\left(\frac{0-{\left(24.07\text{m/s}\right)}^2}{2\left(-9.8{\text{m/s}}^{\text{2}}\right)}\right)\\ =7.0\text{m}+29.09\text{m}\\ \text{=36}.09\text{m}\end{array}$

Therefore, the cannonball reaches the maximum height above the ground is $\underset{\_}{36.09\text{m}}$.

(b)

To determine

The horizontal range of the cannon ball.

Answer to Problem 51P

The horizontal range of the cannonball is $156.94\text{m}$.

Explanation of Solution

Write the expression for the horizontal range of the cannonball.

    $R=\frac{v_0^2\sin \left(2\theta \right)}{g}$        (IV)

Here, $R$ is the horizontal range of the cannon ball.

Conclusion:

Substitute ${37}^{\circ }$ for $\theta$, $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $40\text{m/s}$ for $v_0$ in above equation.

    $\begin{array}{c}R=\frac{{\left(40\text{m/s}\right)}^2\sin \left(2\left({37}^{\circ }\right)\right)}{9.8{\text{m/s}}^{\text{2}}}\\ =156.94\text{m}\end{array}$

Therefore, the horizontal range of the cannonball is $156.94\text{m}$.

(c)

To determine

The $x$ and $y$ components of the cannonball’s velocity.

Answer to Problem 51P

The $x$ and $y$ components of the cannonball’s velocity are $31.95\text{m/s}\text{\&}35.87\text{m/s}$.

Explanation of Solution

Write the expression for the $x$ component of the cannonball’s velocity.

    $v_{x0}=v_0\cos \theta$

Here, $v_{x0}$ is the horizontal initial velocity.

Write the expression for the final vertical velocity of the cannonball.

    $v_{yf}^2=v_{y0}^2+2g{h}_{\max }$        (V)

Here, $v_{yf}$ is the final vertical velocity of the cannonball.

Conclusion:

Substitute $40\text{m/s}$ for $v_0$ and ${37}^{\circ }$ for $\theta$ in (I) from part (a),

  $\begin{array}{c}{v}_{x0}=\left(40\text{m/s}\right)\cos {37}^{\circ }\\ =31.95\text{m/s}\end{array}$

Substitute $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $24.07\text{m/s}$ for $v_{y0}$ and $36.09\text{m}$ for $h_{\max }$ in (V).

    $\begin{array}{c}{v}_{yf}^2={\left(24.07\text{m/s}\right)}^2+2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(36.09\text{m}\right)\\ v_{yf}=\sqrt{1286.7289}\text{m/s}\\ =35.87\text{m/s}\end{array}$

Therefore, the $x$ and $y$ components of the cannonball’s velocity are $31.95\text{m/s}\text{\&}35.87\text{m/s}$.