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

we know that,

$x=v_{x}t$

$V_x=V\cos 鈦�胃$

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

Where, $胃$is the angle made by trajectory vector

Now apply the equation of motion:

$y=v_{x}t鈭�\frac{1}{2}g{t}^2$

role="math" $y=(V\sin 鈦�胃)\left(t\right)鈭�\frac{1}{2}g{t}^2$

$y=y\sin 鈦�胃脳\frac{x}{{\displaystyle v}\cos 鈦�胃}鈭�\frac{g}{2}{\left(\frac{x}{v_x}\right)}^2$

$y\left(x\right)=x\tan 鈦�胃鈭�\frac{1}{2}\frac{g{x}^2}{v_x^2}$

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

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

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.

The rocket's trajectory is a straight line.

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

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,$胃={44.7}^{鈭�}$

Calculate the net force in the upward direction:

$F=T\sin 鈦�胃鈭�mg$

role="math" localid="1648356762974" $=140700\sin {44.7}^{鈭�}鈭�5000脳9.81$

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

Calculate the net force in the horizontal direction:

$F_x=140700脳\cos 44.7$

$F_x=100009.48N$

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

$F=111797.40N$

Calculate the acceleration in the upward direction:

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

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

Using the law of equation of motion:

$v^2鈭�u^2=2as$

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" $=\frac{{330}^2}{2脳22.36}$

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

$Tan胃=\frac{F_y}{F_x}$

$Tan胃=\frac{49967.63}{10009.48}$

$Tan胃=0.4996289$

$胃=26.{55}^0$

Next find the elevation:

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

$h=1088.459$m

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