How High Will a Failing Rocket Reach?

[QuarkCharmer]

Homework Statement

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s^2 and feels no appreciable air resistance. When it has reached a height of 570 m, its engines suddenly fail so that the only force acting on it is now gravity

a.)What is the maximum height this rocket will reach above the launch pad?

b.)How much time after engine failure will elapse before the rocket comes crashing down to the launch pad?

c.)How fast will it be moving just before it crashes?

Homework Equations

Vertical movement equations:
$\Delta x=v_{i}t + \frac{at^{2}}{2}$
$v_{f} = v_{i} + at$
$v_{f}^{2} = v_{i}^{2} = 2a \Delta x$

The Attempt at a Solution

I think the 7600kg is irrelevant, we have not gone over any equations that take mass into account. I'm not sure how to set this problem up.

 

[QuarkCharmer]

I'm using the projectile motion formula:
$$\Delta y = v_{i}sin(\alpha)t +\frac{1}{2}(-9.8)t^{2}$$

But since it's firing straight up, I let alpha be 90 degrees and so:

570 = 2.35t + \frac{1}{2}(-9.8)t^{2}

How can I get it's max height out of this?

 

[lewando]

Since the acceleration is not constant over the entire flight, you cannot use a constant acceleration equation for the whole thing. You must break the problem into two parts. Each with constant acceleration.

 

[Chrisas]

I'm using the projectile motion formula:
$$\Delta y = v_{i}sin(\alpha)t +\frac{1}{2}(-9.8)t^{2}$$

But since it's firing straight up, I let alpha be 90 degrees and so:

570 = 2.35t + \frac{1}{2}(-9.8)t^{2}

How can I get it's max height out of this?

2.35 is given as the upward acceleration, it is not the initial velocity like you are trying to use it as. You are using gravity for acceleration which is not correct for the upward motion when the engines are firing.

Added: The rocket is starting from rest on the launchpad.