Chapter 11.4, Problem 11.112P

To determine

(a)

The angle $\alpha$ that $v_0$ forms with the vertical.

Answer to Problem 11.112P

$\alpha =4.167°$

Explanation of Solution

Given information:

Initial velocity $v_0=75m/s$

Rocket lands at distance $d=100m$ from point A

For a uniformly accelerated motion,

$x=x_0+v_{0}t+\frac{1}{2}a{t}^2$

In above equation,

$x$ - End distance

$x_0$ - Start distance

$v_0$ - Initial velocity

$t$ - Elapsed time

$a$ -Acceleration

Calculation:

For horizontal motion,

$\begin{array}{l}x=x_0+v_{0}t+\frac{1}{2}a{t}^2\\ x=0+v_0\sin \alpha t+0\\ x=v_0\sin \alpha t\end{array}$

Therefore,

$\sin \alpha =\frac{x}{v_{0}t}$

For vertical motion,

$\begin{array}{l}y=y_0+v_{0}t+\frac{1}{2}a{t}^2\\ y=v_0\cos \alpha t+\frac{1}{2}\left(-g\right)t^2\end{array}$

Therefore,

$\cos \alpha =\frac{y+\frac{1}{2}g{t}^2}{v_{0}t}$

But, we know that,

${\sin }^2\alpha +{\cos }^2\alpha =1$

Substitute,

${\left(\frac{1}{v_{0}t}\right)}^2\left(x^2+{\left(y+\frac{1}{2}g{t}^2\right)}^2\right)=1$

Solve further,

$\begin{array}{l}{x}^2+y^2+gy{t}^2+\frac{1}{4}{g}^2{t}^4=v_0{}^2{t}^2\\ \frac{1}{4}{g}^2{t}^4-\left(v_0{}^2-gy\right)t^2+\left(x^2+y^2\right)=0\end{array}$

But,

$\begin{array}{l}\sqrt{x^2+y^2}=100m\\ x=100\cos 30°\\ y=-100\sin 30°=-50m\end{array}$

Therefore,

$\begin{array}{l}\frac{1}{4}{\left(9.81\right)}^2{t}^4-\left({75}^2-\left(9.81\right)\left(-50\right)\right)t^2+{100}^2=0\\ 24.059t^4-6115.5t^2+10000=0\end{array}$

Solve to find $t$

$\begin{array}{l}{t}^2=252.54s^2\\ t^2=1.6458s^2\end{array}$

Therefore,

$\begin{array}{l}t=15.891s\\ t=1.282s\end{array}$

To find the angle $\alpha$

$\tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{x}{y+\frac{1}{2}g{t}^2}=\frac{100\cos 30°}{-50m+\frac{1}{2}\left(9.81\right){\left(15.891\right)}^2}$

Therefore,

$\alpha =4.167°$

Conclusion:

The value of angle $\alpha$ is equal to is calculated using trigonometric equation.

To determine

(b)

The maximum height reached by the rocket?

Answer to Problem 11.112P

$y_{max}=285.18m$

Explanation of Solution

Given information:

Initial velocity $v_0=75m/s$

Rocket lands at distance $d=100m$ from point A

For a uniformly accelerated motion,

$x=x_0+v_{0}t+\frac{1}{2}a{t}^2$

In above equation,

$x$ - End distance

$x_0$ - Start distance

$v_0$ - Initial velocity

$t$ - Elapsed time

$a$ -Acceleration

Calculation:

At maximum height,

$v_y=0$

Therefore,

$\begin{array}{l}{v}_y=v_{y_0}+at\\ 0=v_0\cos \alpha -gt\end{array}$

We get,

$t=\frac{v_0\cos \alpha }{g}$

To find maximum height,

$\begin{array}{l}y=y_0+v_{0}t+\frac{1}{2}a{t}^2\\ y_{\max }=0+v_0\cos \alpha \left(\frac{v_0\cos \alpha }{g}\right)+\frac{1}{2}\left(-g\right){\left(\frac{v_0\cos \alpha }{g}\right)}^2\\ y_{max}=\frac{v_0{}^2{\cos }^2\alpha }{2g}\end{array}$

Substitute,

$y_{max}=\frac{{\left(75m/s\right)}^2{\cos }^2\left(4.167°\right)}{2\left(9.81m/s^2\right)}=285.18m$

Conclusion:

The maximum height reached by the rocket is equal to $y_{max}=285.18m$

To determine

(c)

The duration of flight?

Answer to Problem 11.112P

$t=15.891s$

Explanation of Solution

Given information:

Initial velocity $v_0=75m/s$

Rocket lands at distance $d=100m$ from point A

For a uniformly accelerated motion,

$x=x_0+v_{0}t+\frac{1}{2}a{t}^2$

In above equation,

$x$ - End distance

$x_0$ - Start distance

$v_0$ - Initial velocity

$t$ - Elapsed time

$a$ -Acceleration

Calculation:

For horizontal motion,

$\begin{array}{l}x=x_0+v_{0}t+\frac{1}{2}a{t}^2\\ x=0+v_0\sin \alpha t+0\\ x=v_0\sin \alpha t\end{array}$

Therefore,

$\sin \alpha =\frac{x}{v_{0}t}$

For vertical motion,

$\begin{array}{l}y=y_0+v_{0}t+\frac{1}{2}a{t}^2\\ y=v_0\cos \alpha t+\frac{1}{2}\left(-g\right)t^2\end{array}$

Therefore,

$\cos \alpha =\frac{y+\frac{1}{2}g{t}^2}{v_{0}t}$

But, we know that,

${\sin }^2\alpha +{\cos }^2\alpha =1$

Substitute,

${\left(\frac{1}{v_{0}t}\right)}^2\left(x^2+{\left(y+\frac{1}{2}g{t}^2\right)}^2\right)=1$

Solve further,

$\begin{array}{l}{x}^2+y^2+gy{t}^2+\frac{1}{4}{g}^2{t}^4=v_0{}^2{t}^2\\ \frac{1}{4}{g}^2{t}^4-\left(v_0{}^2-gy\right)t^2+\left(x^2+y^2\right)=0\end{array}$

But,

$\begin{array}{l}\sqrt{x^2+y^2}=100m\\ x=100\cos 30°\\ y=-100\sin 30°=-50m\end{array}$

Therefore,

$\begin{array}{l}\frac{1}{4}{\left(9.81\right)}^2{t}^4-\left({75}^2-\left(9.81\right)\left(-50\right)\right)t^2+{100}^2=0\\ 24.059t^4-6115.5t^2+10000=0\end{array}$

Solve to find $t$

$\begin{array}{l}{t}^2=252.54s^2\\ t^2=1.6458s^2\end{array}$

Therefore,

$\begin{array}{l}t=15.891s\\ t=1.282s\end{array}$

Conclusion:

The possible duration of flight is equal to $t=15.891s$