# Gravitation energy

1. Jun 21, 2014

### Karol

1. The problem statement, all variables and given/known data
A rocket is shot with velocity smaller than the escape velocity. at what angle to the horizon should it be fired in order to reach maximum height.

2. Relevant equations
Total energy: $E=\frac{1}{2}mv^2-\frac{GMm}{r}$

3. The attempt at a solution
The equation isn't vectorial, it doesn't say anything about direction, so, i guess in any direction you shoot it it will reach the same height, but the answer is 900

Last edited: Jun 21, 2014
2. Jun 21, 2014

### PhysicoRaj

Greater than? Lesser than?

3. Jun 21, 2014

### Karol

I fixed, smaller than

4. Jun 21, 2014

### PhysicoRaj

With the velocity less than escape velocity, the rocket executes projectile motion. So when is the height maximum?