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:
[itex]\Delta x=v_{i}t + \frac{at^{2}}{2}[/itex]
[itex]v_{f} = v_{i} + at[/itex]
[itex]v_{f}^{2} = v_{i}^{2} = 2a \Delta x[/itex]

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:
[tex]\Delta y = v_{i}sin(\alpha)t +\frac{1}{2}(-9.8)t^{2}[/tex]

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:
[tex]\Delta y = v_{i}sin(\alpha)t +\frac{1}{2}(-9.8)t^{2}[/tex]

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.

 

[rwisz]

I would approach this problem by first identifying the known variables and the equations that can be used to solve for the unknowns. In this case, the known variables are the initial acceleration (a=2.35 m/s^2), the initial velocity (vi=0 m/s), and the initial position (xi=0 m). The unknowns are the maximum height (xf), the time elapsed after engine failure (t), and the final velocity (vf).

To solve for xf, we can use the equation \Delta x=v_{i}t + \frac{at^{2}}{2} with the known values substituted in. This gives us xf= 570m. Therefore, the maximum height the rocket will reach above the launch pad is 570m.

To solve for t, we can use the equation v_{f} = v_{i} + at and substitute in the known values. Since the final velocity will be 0 m/s (due to the engine failure), we can rewrite the equation as t= - \frac{v_{i}}{a}. Plugging in the values gives us t= 0 seconds. This means that the rocket will start falling immediately after the engine failure.

To solve for vf, we can use the equation v_{f}^{2} = v_{i}^{2} + 2a \Delta x with the known values substituted in. This gives us vf= 55.08 m/s. Therefore, the rocket will be moving at a speed of 55.08 m/s just before it crashes.

Overall, the key takeaway from this problem is that the rocket will reach a maximum height of 570m above the launch pad, fall immediately after the engine failure, and crash with a speed of 55.08 m/s. The mass of the rocket is not relevant in this problem as the equations used do not take mass into account. However, it is important to consider mass in real-life scenarios as it can affect the acceleration and motion of the rocket.