# Rocket Height-Kinematic Question

1. Sep 14, 2009

### Chandasouk

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

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant net acceleration a, until time t1, when the fuel is exhausted.

Find the maximum height H that the rocket reaches (neglecting air resistance).

Express the maximum height in terms of a, t1, and/or g. Note that in this problem, g is a positive number equal to the magnitude of the acceleration due to gravity.

3. The attempt at a solution

Okay, in this problem, i am pretty much stuck. I only did the hints that Mastering Physics suggested I do to help me solve this problem.

I found the height, H1, above the ground at which the rocket exhausted its fuel

H1=1/2at1^2

I then found the y velocity, Vy1, that the rocket has when the engine runs out of fuel

Vy1=at1

Where do I go from here?

2. Sep 14, 2009

### pgardn

So you got the initial velocity when the rocket runs out of fuel (stops accelerating up) so it is now in free fall. What is the velocity of the rocket when it reaches its max height? Then use the kinematic equation that does not involve time but has displacement in the equation. Essentially the last part of the trip is just like throwing a rock up in the air. You got the velocity upon release, and at max height the velocity is zero.

3. Sep 14, 2009

### Staff: Mentor

Once the rocket runs out of fuel it can be treated like any other projectile in freefall. You have its speed when it runs out of fuel, so figure out how much further it rises.

4. Sep 14, 2009

### Chandasouk

Okay, the equation which pgardn suggested is

Vf^2 = Vi^2 + 2ad

The velocity of the rocket at the max height of it's path is 0m/s

Acceleration during free fall equals -9.81m/s^2

I am assuming the Final Velocity (Vf) is 0m/s, so I set it up like this

(0m/s)^2 = Vi^2 + 2ad

Previously, I found the Y velocity which was Vy1 = at1. I use this as a substitute

(0m/s)^2 = (at1)^2 + 2(-9.81m/s^2)d

0-(at1)^2=-19.62m/s^2*d

-(at1)^2
------------- = D ?
-19.62m/s^2

In terms of a, t1, and/or g, it would be

H = -(at)^2
-------- ?
- 2g

5. Sep 15, 2009

### Staff: Mentor

That's the additional height the rocket will reach measured from the point where it ran out of fuel. What's the total height above the ground? (And please cancel those minus signs.)

6. Sep 15, 2009

### Chandasouk

So, previously, I found the height, H1, above the ground at which the rocket exhausted its fuel

H1=1/2at1^2

and also using my last equation

H = (at)^2
--------
2g

The total distance covered by the rocket would equal the sum of these two equations

H=1/2at1^2 + (at1^2/2g) ?

7. Sep 15, 2009

### Staff: Mentor

Looks good to me.

8. Sep 15, 2009

### Chandasouk

Mastering Physics told me this answer was incorrect because

9. Sep 15, 2009

### rl.bhat

It should be
H = 1/2*at12 +1/2 (at2)2/g

10. Sep 15, 2009

### Staff: Mentor

You misplaced your parentheses. It should be:
H=1/2at1^2 + (at1)^2/2g

(Sorry for not catching that earlier.)

Use subscripts/superscripts for further clarity:
H=1/2at12 + (at1)2/2g