Homework Help: Maximum altitude of a rocket launch

1. Sep 1, 2010

imatreyu

1. The problem statement, all variables and given/known data

A rocket is launched at an angle of 53 degrees above the horizontal with an intial speed of 100 m/s. It moves along its initial line of motion with an accel. of 30 m/s^2 for 3 s. At this time its engines fail and the rocket proceeds to move as a free body. Find the maximum altitude reached by the rocket.

2. Relevant equations
Just the four basic kinematics equations. . ?

3. The attempt at a solution
I did:

where is the rocket when its engines fail?
d= vit + (1/2)at^2
= 100 sin 53 (3) + (1/2)(30)(3^2)
= 374.59 m

what is the rocket's speed after the engines fail?
vf= vi + at
=100 + 30 (3)
=190 m/s

So I feel like from here I should just be able to plug into vf^2= vi^2 + 2ad and find "d," using vi= 190 m/s and vf= 0. . . .But that's not the case. Why can't I do that? And where can I go from here? I have a feeling I need to find some way to express time to reach the maximum altitude, but I have no idea how I can express that.

2. Sep 1, 2010

Gyro

At what angle is the rocket travelling when its engines cut?

3. Sep 1, 2010

imatreyu

The rocket is traveling at 53 degrees above the horizontal when its engine is cut.

4. Sep 1, 2010

Gyro

So what are the components of its velocity at the time that the engine cuts? How can you use the vertical component to calculate the time it takes to reach maximum altitude? How can that time be used to find the vertical displacement of the rocket after the engines cut?

5. Sep 1, 2010

imatreyu

1. Components of initial velocity: vix= 100cos53, viy= 100sin53
2. I honestly have no idea. :( I don't know what equation to create. . how to express time. . .
3. . . .

6. Sep 1, 2010

Gyro

When the engine fails, it's velocity is 190 m/s at 53 degrees from horizontal.

Then in the vertical direction, you can use the equation Vf = Vi + at, where Vf = 0, Vi = 190sin53 m/s , a = -9.8 m/s^2, and isolate t.

Go from there.

7. Sep 1, 2010

imatreyu

Oh. . .
So viy= 190sin53, vfy= 0?

I think I can see how I could use that to find time to fall in one of the kinematics equations, but I don't see how I could express time to reach maximum altitude.

8. Sep 1, 2010

Gyro

Remeber that after engines fails, the only acceleration on the rocket is gravity.

9. Sep 1, 2010

imatreyu

Is time even relevant, then?

Or can I just do this:

vf^2 = vi^2 -2(9.8)d
vf= 0 (because at apogee velocity is s 0, and apogee would be the maximum altitude)
viy= 190sin53

and then solve for d?

10. Sep 1, 2010

Gyro

Yes, but breaking it down helps to visualize it, no? And you asked for a way to express time, so I showed you.

Plus, it asks for max altitude, this 'd' is only from free fall. We have to add the 'd' from rocket thrust upwards.

Hope I helped.

Last edited: Sep 1, 2010
11. Sep 1, 2010

imatreyu

Ohh. . . .Haha the time thing was a result of Yahoo answers confusion.

Oh, and then I add the initial distance.
Aha! I get it!!!
THANK YOU SO MUCH!!!