Chapter 3, Problem 4SP

(a)

To determine

The time taken by the cannonball to remain in the air, if it is fired at an angle of $20°$ so that the vertical component of velocity is $137\text{m}/\text{s}$ and horizontal component of initial velocity is $376\text{m}/\text{s}$.

Answer to Problem 4SP

The time taken by the cannonball to remain in air is $27.4\text{s}$.

Explanation of Solution

Given info: The vertical component of initial velocity is $137\text{m}/\text{s}$ and the acceleration due to gravity at the place is $10\text{m}/{\text{s}}^{\text{2}}$.

Write the equation of motion along the vertical direction.

$v_{\text{vertical}}=v_{\text{0vertical}}+a_{\text{vertical}}t$

Here,

$v_{\text{vertical}}$ is the vertical component velocity

$v_{0\text{vertical}}$ is the vertical component of initial velocity

$a_{\text{vertical}}$ is the acceleration due to gravity

$t$ is the time taken

When the ball reaches the maximum height the velocity becomes zero. Since in the upward direction the velocity is decreases towards zero, the acceleration due to gravity is negative during the upward motion.

Therefore, Substitute $137\text{m}/\text{s}$ for $v_{0\text{vertical}}$ , $0\text{m}/\text{s}$ for $v_{\text{vertical}}$ and $-10{\text{m}/\text{s}}^2$ for $a_{\text{vertical}}$ in the above equation to get .

$\begin{array}{c}0\text{m}/\text{s}=137\text{m}/\text{s}+\left(-10\text{m}/{\text{s}}^2\right)t\\ t=\frac{-137\text{m}/\text{s}}{-10\text{m}/{\text{s}}^2}\\ =13.7\text{s}\end{array}$

This is the time taken by the ball to reach the maximum height.

Since total time of flight is twice the time required to reach the high point, the actual time taken by the ball to remain in air is equal to $2\times 13.7\text{s}\text{=}\text{27}\text{.4}\text{s}$.

Conclusion:

Thus, total time taken by the ball to remain in the air is $27.4\text{s}$.

(b)

To determine

The horizontal distance travelled by the cannonball, if it is fired at an angle of $20°$ so that the vertical component of velocity is $137\text{m}/\text{s}$ and horizontal component of initial velocity is $376\text{m}/\text{s}$.

Answer to Problem 4SP

The horizontal distance travelled by the cannonball is $10,300\text{m}$.

Explanation of Solution

Given info: The horizontal component of initial velocity is $376\text{m}/\text{s}$.

In the horizontal direction the there is no acceleration, the ball is travelling at constant velocity.

Write the equation of motion along the vertical direction.

$d_{\text{horizontal}}=v_{\text{0horizontal}}t+\frac{1}{2}{a}_{\text{horizontal}}{t}^2$

Here,

$d_{\text{horizontal}}$ is the horizontal distance travelled by the cannonball

$v_{\text{horizontal}}$ is the vertical component velocity

$v_{0\text{horizontal}}$ is the vertical component of initial velocity

$a_{\text{horizontal}}$ is horizontal component of acceleration

$t$ is the time taken

The time taken by the cannonball to remain in air is $27.4\text{s}$.

Since there is no acceleration in the vertical direction, put $0{\text{m}/\text{s}}^2$ for $a$ in the above equation to get $d_{\text{horizontal}}$.

$d_{\text{horizontal}}=v_{\text{0horizontal}}t$

Substitute $376\text{m}/\text{s}$ for $v_{\text{0horizontal}}$ and $27.4\text{s}$ for $t$ in the above equation to get $d_{\text{horizontal}}$.

$\begin{array}{c}{d}_{\text{horizontal}}=\left(376\text{m}/\text{s}\right)\left(27.4\text{s}\right)\\ =10,300\text{m}\end{array}$

Conclusion:

Thus, the horizontal distance travelled by the cannonball is $10,300\text{m}$.

(c)

To determine

The time taken by the cannonball to remain in the air and the horizontal distance travelled by the cannonball, if it is fired at an angle of $70°$ so that the vertical component of velocity is $376\text{m}/\text{s}$ and horizontal component of initial velocity is $137\text{m}/\text{s}$.

Answer to Problem 4SP

The time taken by the cannonball to remain in air is $75.2\text{s}$ and the horizontal distance travelled by the cannonball is $10,300\text{m}$, when the angle of projection is $20°$, which is same as the horizontal distance travelled by the cannonball fired at an angle of $70°$.

Explanation of Solution

If the vertical component of velocity is $376\text{m}/\text{s}$ and horizontal component of initial velocity is $137\text{m}/\text{s}$.

Write the equation of motion along the vertical direction.

$v_{\text{vertical}}=v_{\text{0vertical}}+a_{\text{vertical}}t$

Here,

$v_{\text{vertical}}$ is the vertical component velocity

$v_{0\text{vertical}}$ is the vertical component of initial velocity

$a_{\text{vertical}}$ is the acceleration due to gravity

$t$ is the time taken

When the ball reaches the maximum height the velocity becomes zero. Since in the upward direction the velocity is decreases towards zero, the acceleration due to gravity is negative during the upward motion.

Therefore, Substitute $376\text{m}/\text{s}$ for $v_{0\text{vertical}}$ , $0\text{m}/\text{s}$ for $v_{\text{vertical}}$ and $-10{\text{m}/\text{s}}^2$ for $a_{\text{vertical}}$ in the above equation to get .

$\begin{array}{c}0\text{m}/\text{s}=376\text{m}/\text{s}+\left(-10\text{m}/{\text{s}}^2\right)t\\ t=\frac{-376\text{m}/\text{s}}{-10\text{m}/{\text{s}}^2}\\ =37.6\text{s}\end{array}$

This is the time taken by the ball to reach the maximum height.

Since total time of flight is twice the time required to reach the high point, the actual time taken by the ball to remain in air is equal to $2\times 37.6\text{s}\text{=}\text{75}\text{.2}\text{s}$.

Write the equation of motion along the vertical direction.

$d_{\text{horizontal}}=v_{\text{0horizontal}}t+\frac{1}{2}{a}_{\text{horizontal}}{t}^2$

Here,

$d_{\text{horizontal}}$ is the horizontal distance travelled by the cannonball

$v_{\text{horizontal}}$ is the vertical component velocity

$v_{0\text{horizontal}}$ is the vertical component of initial velocity

$a_{\text{horizontal}}$ is horizontal component of acceleration

$t$ is the time taken

The time taken by the cannonball to remain in air is $\text{75}\text{.2}\text{s}$.

Since there is no acceleration in the vertical direction, put $0{\text{m}/\text{s}}^2$ for $a$ in the above equation to get $d_{\text{horizontal}}$.

$d_{\text{horizontal}}=v_{\text{0horizontal}}t$

Substitute $137\text{m}/\text{s}$ for $v_{\text{0horizontal}}$ and $\text{75}\text{.2}\text{s}$ for $t$ in the above equation to get $d_{\text{horizontal}}$.

$\begin{array}{c}{d}_{\text{horizontal}}=\left(137\text{m}/\text{s}\right)\left(\text{75}\text{.2}\text{s}\right)\\ =10,300\text{m}\end{array}$

Conclusion:

Thus, time taken by the cannonball to remain in air is $75.2\text{s}$ and the horizontal distance travelled by the cannonball is $10,300\text{m}$, when the angle of projection is $20°$, which is same as the horizontal distance travelled by the cannonball fired at an angle of $70°$.