Answer to Question #132857 in Mechanics | Relativity for M

Question

A rocket is launched at an angle of 70° above the horizontal from rest. It moves for 30s along this path with an acceleration of 45m/s2. After this time, the engines fail and the rocket proceeds to move as a free body.

a. What is the rocket’s maximum altitude?

b. What is the rocket’s total flight time?

c. What is the rocket’s impact distance (horizontal range)?

Answer 1

The first thing we have to do is to get the corresponding acceleration in X axis and Y axis using trigonometric formula. Once we have the Y component of acceleration, we subtract it with the acceleration due to gravity to get the net acceleration. We then use the formulas. So to understand more about the process solving this kind of problems.

Explanation

$Q_1$ Rocket maximum altitude.

. $Y=Vot+ \frac{1}{2}at^{2}$

$a=aY-ag$

$a=92.29m/s^{2}-9.8m/s^{2}$

$a=82.48m/s^2$

$t=30 sec$

$Y=Vot+\frac{1}{2}\times82.48 \times 30^{2}$

$Y=14,616 meters$

$Q_2$ Rocket total flight

$Y=Vot+\frac{1}{2}at^{2}$

$-14616=Vot+\frac{1}{2}\times-9.8\times t^{2}$

$-14616=-4.90t^{2}$

$t=√2979.82$

$t=54.59 sec$

Total flight time = 30 sec (going up)+54.59sec going down

$=84.59 seconds$

$Q_3$ Rockets impact distance

$X=Vot+\frac{1}{2}at^{2}$

$a=aX=15.39m/s;V_0=0;t=84.59$

$X=\frac{1}{2}\times15.39\times84.59^2$

$X=55,058.33 meters$