Question

A rocket is launched from rest and moves in a straight line at 30.0 degrees above the horizontal with an acceleration of 35.0 m/s2. After 25.0 s of powered flight, the engines shut off and the rocket follows a parabolic path back to earth. Find the time of flight from launch to impact. HINT: Simple projectile motion after engines are shut down.

Answer 1

Answer:

t = 123.59s

Explanation:

For the launch pad section:

Vf = Vo + a*t  where Vo=0.

Vf = 35*25 = 875m/s

The distance traveled during the launch:

$d = Vo*t+\frac{a*t^2}{2} = 0+\frac{35*25^2}{2} = 10937.5m$

Now the projectile motion, we know that its initial speed is the speed calculated previously and the initial height is the y-component of the previously calculated distance.

$\Delta Y = Vo*sin(30)*t - \frac{g*t^2}{2}$

$-d*sin(30) = Vo*sin(30)*t - \frac{g*t^2}{2}$  where d= 10937.5m; Vo=875m/s.

Solving for t:

t1 = -11.093s   t2 = 98.59s

So, the total time of flight will be:

$t_{total} = t_{launch}+t_{projectile}=25+98.59 = 123.59s$