A 7800kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25m/s^2 and feels no appreciable air resistance. When it has reached a height of 540m , its engines suddenly fail so that the only force acting on it is now gravity. What is the maximum height this rocket will reach above the launch pad?

Question

noelgreco · Answer

First, figure out how fast the rocket is going when the engines fail.

$$v _{f}^{2}=v _{i}^{2}+2ad$$

anonymous · Answer

what would be the initial velocity? it doesn't say the rocket starts from rest?

anonymous · Answer

@stokestheorem Are you going to use only kinematics?  Or is energy also allowed?

anonymous · Answer

Start with $t=0, x_i=0, v_i=0$.  Try to find $t$ when $x(t)=h$.

anonymous · Answer

You're going to have constant acceleration.  
There will be $a_e$ when engines are working and $a_g$ when they have failed and you're in free fall.  The position from time function until engine failure is $$
x(t) = \frac{1}{2}a_et^2+v_it+x_i
$$

anonymous · Answer

@wio kinematics thanks for answering i'll try ur suggestins out

anonymous · Answer

Once you have found the time.  You create a new parabola due to our change in acceleration.  Where $v_i = a_et$ and $x_i=h$ $$
x(t) = \frac{1}{2}a_gt^2+v_it+x_i
$$

anonymous · Answer

@wio i dont quite understand the point of setting xi, t , and xi =0

anonymous · Answer

The point is to solve the problem: "What time/velocity are we when the engine fails"

anonymous · Answer

Once we get that velocity, we can move on to the second part of the problem.

anonymous · Answer

the velocity is 0m/s at the max height, but I dont know how long it took to get up there from the launch pad

anonymous · Answer

But it is NOT at its max height when the engine fails, get it?

anonymous · Answer

Do you want me to just do it for you as an example?

anonymous · Answer

@wio thanks for the help; I just needed to understand the problem more