Chapter 3, Problem 4SP

A cannon is fired over level ground at an angle of 35° to the horizontal. The initial velocity of the cannonball is 500 m/s, but because the cannon is fired at an angle, the vertical component of the velocity is 287 m/s and the horizontal component is 410 m/s.

1. a. How long is the cannonball in the air? (Use g = 10 m/s² and the fact that the total time of flight is twice the time required to reach the high point.)
2. b. How far does the cannonball travel horizontally?
3. c. Repeat these calculations, assuming that the cannon was fired at a 55° angle to the horizontal, resulting in a vertical component of velocity of 410 m/s and a horizontal component of 287 m/s. How does the distance traveled compare to the earlier result? (See fig. 3.20.)

To determine

The time of flight of the ball.

Answer to Problem 4SP

The time of flight of the ball is $114.8\text{s}$.

Explanation of Solution

Given Info: A cannon is fired over the ground level at an angle of $35°$ to the horizontal. The initial velocity is $500\text{ m/s}$. The vertical component is $287\text{m/s}$ and the horizontal component is $410\text{m/s}$.

Write the expression for the time required to reach the high point.

$t=\sqrt{\frac{2v_{v0}}{g}}$

Here,

$t$ is the time

$v_{v0}$ is the initial vertical velocity

$g$ is the acceleration due to gravity

Substitute $287\text{m/s}$ for $v_{v0}$ and $10{\text{m/s}}^2$ for to get $t$.

$\begin{array}{c}t=\sqrt{\frac{2\left(287\text{m/s}\right)}{10{\text{m/s}}^2}}\\ =57.4\text{s}\end{array}$

Write the expression for the time of flight.

$T=2t$

Here,

$T$ is the time of flight

Substitute $57.4\text{s}$ for $t$ to get $T$.

$\begin{array}{c}T=2\left(57.4\text{s}\right)\\ =114.8\text{s}\end{array}$

Conclusion:

Thus, the time of flight of the ball is $114.8\text{s}$.

To determine

The distance travelled by the cannonball horizontally.

Answer to Problem 4SP

The distance travelled by the cannonball horizontally is $112.963\text{km}$.

Explanation of Solution

Given Info: A cannon is fired over the ground level at an angle of $35°$ to the horizontal. The initial velocity is $500\text{ m/s}$. The vertical component is $287\text{m/s}$ and the horizontal component is $410\text{m/s}$.

Write the expression for the distance travelled horizontally.

$d_h=v_{0h}t+\frac{1}{2}g{t}^2$

Here,

$d_h$ is the distance travelled

$v_{0h}$ is the initial horizontal velocity

Substitute $410\text{m/s}$ for $v_{0h}$, $10{\text{m/s}}^2$ for $g$ and $114.8\text{s}$ for $t$ to get $d_h$.

$\begin{array}{c}{d}_h=\left(410\text{m/s}\right)\left(114.8\text{s}\right)+\frac{1}{2}\left(10{\text{m/s}}^2\right){\left(114.8\text{s}\right)}^2\\ =112.963\times {10}^3\text{m}\times \frac{1\text{km}}{{10}^3\text{m}}\\ =112.963\text{km}\end{array}$

Conclusion:

Thus, the distance travelled by the cannonball horizontally is $112.963\text{km}$.

To determine

Repeat the above results for angle $55°$.

Answer to Problem 4SP

The time of flight of the ball is $164\text{s}$ and the distance travelled by the cannonball horizontally is $181.548\text{km}$.

Explanation of Solution

Given Info: A cannon is fired over the ground level at an angle of $55°$ to the horizontal. The initial velocity is $500\text{ m/s}$. The vertical component is $410\text{m/s}$ and the horizontal component is $287\text{m/s}$.

Write the expression for the time required to reach the high point.

$t=\sqrt{\frac{2v_{v0}}{g}}$

Substitute $410\text{m/s}$ for $v_{v0}$ and $10{\text{m/s}}^2$ for to get $t$.

$\begin{array}{c}t=\sqrt{\frac{2\left(410\text{m/s}\right)}{10{\text{m/s}}^2}}\\ =82\text{s}\end{array}$

Write the expression for the time of flight.

$T=2t$

Substitute $82\text{s}$ for $t$ to get $T$.

$\begin{array}{c}T=2\left(82\text{s}\right)\\ =164\text{s}\end{array}$

Write the expression for the distance travelled horizontally.

$d_h=v_{0h}t+\frac{1}{2}g{t}^2$

Substitute $287\text{m/s}$ for $v_{0h}$, $10{\text{m/s}}^2$ for $g$ and $164\text{s}$ for $t$ to get $d_h$.

$\begin{array}{c}{d}_h=\left(287\text{m/s}\right)\left(164\text{s}\right)+\frac{1}{2}\left(10{\text{m/s}}^2\right){\left(164\text{s}\right)}^2\\ =181.548\times {10}^3\text{m}\times \frac{1\text{km}}{{10}^3\text{m}}\\ =181.548\text{km}\end{array}$

Conclusion:

Thus, the time of flight of the ball is $164\text{s}$ and the distance travelled by the cannonball horizontally is $181.548\text{km}$.