A 7500 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s^2?

Question

When it has reached a height of 590 m, 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?
How much time after engine failure will elapse before the rocket comes crashing down to the launch pad?
How fast will it be moving just before it crashes?

Answer 1

Since the rocket has constant acceleration upward to 530 m, we can compute the time to reach that altitude (assuming constant gravity and no loss of mass)
y(t)=590=.5*2.35*t^2
solve for t
t=sqrt(1180/2.35)

Now use that t to compute the velocity of the rocket when the engine fails:
v(t)=2.35*t
plug in t from above
v(t)=sqrt(2.35*1180)

lets call this v0 and 590 m is y0

The next phases of the flight are up to apogee followed by a crash to Earth.

The apogee occurs when v(t)=0, and now
v(t)=v0-g*t
so
v0/g=t

The height is
y(t)=590+v0*t-.5*g*t^2
plugging in t
apogee=590+.5*v0^2/g

The rocket will crash into the launch pad when
y(t)=0
so
0=590+v0*t-.5*g*t^2
solve for the two roots. There will be a positive root and a negative root. The negative root is irrelevant to this problem (although, because of the symmetry of parabolic motion, it is interesting in that if the rocket were launched instantaneously at that time, and at the same speed as the crash, it would follow the same trajectory above 590 m and down).

Take the positive root, that's the time from engine failure to crash.

the speed is found using that t in the following equation
v(t)=v0-g*t

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