Chapter 8.1, Problem 11E

a.

To determine

To calculate velocity of the projectile after 3 seconds.

Answer to Problem 11E

Velocity after 3 seconds is $-6$ feet per second.

Explanation of Solution

Given information:

The acceleration due to Earth’s gravity is 32 ${\text{ft/sec}}^2$ and the projectile is fired straight upward with a velocity of 90 feet per second.

Take velocity and acceleration in the upward direction as positive and in the downward direction as negative.

The time interval is $\left[0,3\right]$

  $\begin{array}{l}\frac{dv}{dt}=a\\ \text{change in velocity}={\displaystyle \underset{0}{\overset{3}{\int }}adt}\\ v_2-v_1=at{]}_0^3\left[v_{2,}{v}_1\text{represents final and initial velocity respectively}\right]\\ v_2-v_1=a\left[3-0\right]\\ v_2-v_1=-32\cdot 3\\ v_2=-96+v_1\\ v_2=-96+90\\ v_2=-6\end{array}$

Therefore,

Velocity after 3 seconds is $-6$ feet per second.

b.

To determine

To calculate when the projectile hits the ground.

Answer to Problem 11E

The projectile hits the ground after 5.625 seconds

Explanation of Solution

Given information:

The acceleration due to Earth’s gravity is 32 ${\text{ft/sec}}^2$ and the projectile is fired straight upward with a velocity of 90 feet per second.

Take velocity and acceleration in the upward direction as positive and in the downward direction as negative.

Velocity of a projectile when it hits the ground is numerically same as velocity with which it is fired but the sign is opposite, therefore velocity of projectile when it hits the ground is $-90$ feet per second.

Consider after x seconds the projectile hits the ground.

  $\begin{array}{l}\frac{dv}{dt}=a\\ \text{change in velocity}={\displaystyle \underset{0}{\overset{x}{\int }}adt}\\ v_2-v_1=at{]}_0^x\left[v_{2,}{v}_1\text{represents final and initial velocity respectively}\right]\\ v_2-v_1=a\left[x-0\right]\\ -90-90=-32\cdot x\left[\text{substitute values of velocity and acceleration}\right]\\ -180=-32x\\ x=\frac{-180}{-32}\left[\text{divide each side by -32}\right]\\ x=5.625\end{array}$

Therefore,

The projectile hits the ground after 5.625 seconds.

c.

To determine

To calculate the net distance travelled when the projectile hits the ground.

Answer to Problem 11E

Net distance travelled is 0 when the projectile hits the ground.

Explanation of Solution

Given information:

The acceleration due to Earth’s gravity is 32 ${\text{ft/sec}}^2$ and the projectile is fired straight upward with a velocity of 90 feet per second.

Take velocity and acceleration in the upward direction as positive and in the downward direction as negative.

Time taken in reaching top $=$ Time taken in reaching ground from top $=\frac{5.625}{2}=2.8125$ seconds.

Net distance travelled $=$ distance travelled in reaching top from ground $+$ distance travelled in reaching ground from top.

  $\begin{array}{l}S={\displaystyle \underset{0}{\overset{2.8125}{\int }}vdt}+{\displaystyle \underset{0}{\overset{2.8125}{\int }}vdt}\\ S={\displaystyle \underset{0}{\overset{2.8125}{\int }}\left(u+at\right)dt}+{\displaystyle \underset{0}{\overset{2.8125}{\int }}\left(u+at\right)dt}\left[u\text{ is the initial velocity}\right]\\ S=ut{]}_0^{2.8125}+\frac{a{t}^2}{2}{]}_0^{2.8125}+\frac{a{t}^2}{2}{]}_0^{2.8125}\\ S=90\cdot 2.8125+\frac{-32\left[{\left(2.8125\right)}^2-0^2\right]}{2}+\frac{-32\left[{\left(2.8125\right)}^2-0^2\right]}{2}\\ S=253.125-32{\left[2.8125\right]}^2\\ S=253.125-253.125\\ S=0\end{array}$

Therefore,

Net distance travelled is 0.

d.

To determine

To calculate total distance travelled by the projectile when it hits the ground.

Answer to Problem 11E

Total distance travelled by the projectile is 253.125 feet when it hits the ground.

Explanation of Solution

Given information:

The acceleration due to Earth’s gravity is 32 ${\text{ft/sec}}^2$ and the projectile is fired straight upward with a velocity of 90 feet per second.

Take velocity and acceleration in the upward direction as positive and in the downward direction as negative.

Time taken in reaching top $=$ Time taken in reaching ground from top $=\frac{5.625}{2}=2.8125$ seconds.

Total distance travelled $=$ distance travelled in reaching top from ground $-$ distance travelled in reaching ground from top.

  $\begin{array}{l}S={\displaystyle \underset{0}{\overset{2.8125}{\int }}vdt}-{\displaystyle \underset{0}{\overset{2.8125}{\int }}vdt}\\ S={\displaystyle \underset{0}{\overset{2.8125}{\int }}\left(u+at\right)dt}-{\displaystyle \underset{0}{\overset{2.8125}{\int }}\left(u+at\right)dt}\left[u\text{ is the initial velocity}\right]\\ S=ut{]}_0^{2.8125}+\frac{a{t}^2}{2}{]}_0^{2.8125}-\frac{a{t}^2}{2}{]}_0^{2.8125}\\ S=90\cdot 2.8125+\frac{-32\left[{\left(2.8125\right)}^2-0^2\right]}{2}-\frac{-32\left[{\left(2.8125\right)}^2-0^2\right]}{2}\\ S=253.125\end{array}$

Therefore,

Total distance travelled by the projectile is 253.125 feet.