Question

A $5000kg$interceptor rocket is launched at an angle of $44.7°$. The thrust of the rocket motor is$140700N$

a. Find an equation $y\left(x\right)$that describes the rocket’s trajectory.

b. What is the shape of the trajectory?

c. At what elevation does the rocket reach the speed of sound, 330 m/s?

Short answer

a). An equation $y\left(x\right)$that describe the rocket's trajectory $y\left(x\right)=x\tan \theta -\frac{1}{2}\frac{g{x}^2}{v{x}^2}$

b). The rocket's trajectory is a straight line.

c). The rocket reach at the elevation of$1088.46m$

Step-by-step solution

Given Infromation (Part a)

A$5000kg$interceptor rocket is launched at an angle of $44.7°$. The thrust of the rocket motor is $140,700N$.

Explanation (Part a)

we know that,

$x=v_{x}t$

$V_x=V\cos \theta$

$V_x=\left(ax\right)\left(t\right)$

Where, $\theta$is the angle made by trajectory vector

Now apply the equation of motion:

$y=v_{x}t-\frac{1}{2}g{t}^2$

role="math" $y=(V\sin \theta )\left(t\right)-\frac{1}{2}g{t}^2$

$y=y\sin \theta \times \frac{x}{{\displaystyle v}\cos \theta }-\frac{g}{2}{\left(\frac{x}{v_x}\right)}^2$

$y\left(x\right)=x\tan \theta -\frac{1}{2}\frac{g{x}^2}{v_x^2}$

Final Answer (Part a)

An equation $y\left(x\right)$that describe the rocket's trajectory is$y\left(x\right)=x\tan \theta -\frac{1}{2}\frac{g{x}^2}{V_x^2}$

Given Information (Part b)

A $5000kg$ interceptor rocket is launched at an angle of $44.7°$. The thrust of the rocket motor is $140,700N$.

Explanation (Part b)

The equation $y\left(x\right)=\frac{x}{2}$ is one of a straight line, where the $y$-intercept is zero. Therefore, the rocket's trajectory is a straight line.

Final Answer (Part b)

The rocket's trajectory is a straight line.

Given Information (Part c)

A $5000kg$ interceptor rocket is launched at an angle of $44.7°$. The thrust of the rocket motor is $140,700N$.

Explanation (Part c)

According to the information:

Mass of the rocket $m=5000\mathrm{kg}$

Thrust of rocket $T=140700\mathrm{N}$

Angle made by the rocket with the horizontal,$\theta ={44.7}^{\circ }$

Calculate the net force in the upward direction:

$F=T\sin \theta -mg$

role="math" localid="1648356762974" $=140700\sin {44.7}^{\circ }-5000\times 9.81$

role="math" localid="1648356783632" $=49967.63\mathrm{N}$

Calculate the net force in the horizontal direction:

$F_x=140700\times \cos 44.7$

$F_x=100009.48N$

$F=\sqrt{100009.{48}^2+49967.{63}^2}$

$F=111797.40N$

Calculate the acceleration in the upward direction:

$\mathrm{a}=\frac{{\mathrm{F}}_{\mathrm{net}}}{\mathrm{m}}$

role="math" localid="1648357149228" $=\frac{11797.40}{5000}$

role="math" localid="1648357176500" $=22.36m/s^2$

Applying the equation of motion

Using the law of equation of motion:

$v^2-u^2=2as$

Where,

$v$is the initial velocity

$u$is the final velocity

$s$is the distance covered

$s=\frac{v^2}{2a}$

role="math" localid="1648357201368" $=\frac{{330}^2}{2\times 22.36}$

role="math" localid="1648357225685" $=2435.15\mathrm{m}$

$Tan\theta =\frac{F_y}{F_x}$

$Tan\theta =\frac{49967.63}{10009.48}$

$Tan\theta =0.4996289$

$\theta =26.{55}^0$

Next find the elevation:

Elevation role="math" localid="1648357435052" $h=2435.15\times \sin \left(26.{55}^0\right)$

$h=1088.459$m

Final Answer 

The rocket reach the speed of sound at the elevation of$1088.46m$