# Speed of a rocket using energy methods

1. May 22, 2014

### BrainMan

1. The problem statement, all variables and given/known data
A rocket is launched such that when the fuel is exhausted, the rocket is moving with a speed of Vo at an angle of 37° with the horizontal and at an altitude h. (a) Use energy methods to find the speed of the rocket when its altitude is h/2. (b) Find the x and y components of velocity when the rockets altitude is h/2. Use the fact that Vx=Vxo = constant (since ax-0) and the result from a.

2. Relevant equations
The Kinetic and Potential energy equations

3. The attempt at a solution
I did
1/2mv2+mgy= E
and then I solved for v and got
v= (2Egy)1/2
the correct answer is (a) v = (v02 + gh)1/2
(b) vx= .799vo, vy= - (- 0.362vo2 +gh)1/2
I wasn't really sure how to approach this type of problem
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 22, 2014

### vela

Staff Emeritus
How did you manage to get that? It looks like you didn't do the algebra correctly.

3. May 22, 2014

### BiGyElLoWhAt

in a you solved for v wrong. You're attempting to do what they did, you just made a mistake that I think you can find.
When they say the speed at h/2, do they mean on the way up or down?

4. May 22, 2014

### BrainMan

I'm not sure and what could I do with that information?

5. May 22, 2014

### haruspex

You have two instances where you can apply that equation: at the point where the fuel is exhausted, and again when it has descended to height h/2. Which term is the same at both points? What equation does that give you?

6. May 22, 2014

### BrainMan

Is it that they both have the same amount of energy? So would you get something like 1/2mv2+mgy=1/2mv2+mgy?

7. May 22, 2014

### haruspex

Yes, plugging in the appropriate values for v and y each side.

8. May 24, 2014

### BrainMan

OK I get it but I am having trouble finding the y component. I did sin 37 to find the y component but it looks like in the answer they gave they have that number squared - (0.362vo2 +gh)1/2. Why is sin theta squared in this equation?

9. May 24, 2014

### SammyS

Staff Emeritus
What is $\ (v_0)_y\ ?\$ Isn't sin(θ0) a factor of this?

10. May 24, 2014

### BrainMan

The initial speed in the y direction is Vo* sin(37). So shouldn't the equation be .601Vo instead of .362vo?

11. May 24, 2014

### haruspex

If a speed is 0.601v, what is the square of that speed?

12. May 24, 2014

### BrainMan

OK I see how they got that answer. Thanks everyone!