# Projectile motion, 30 and 60 degrees on a level plain

1. Oct 18, 2007

### nooobie

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

prove that two bullets fired at 30 and 60 degrees will travel the same distance, and the 60 degree shot will travel 3 times as high as the 30 degree.

2. Relevant equations
S(displacement) Vo(initial velocity) Vf(final Velovity) t(time) a(acceleration due to gravity)

S=Vot + .5at^2
S=Vft - .5at^2
S= (Vft - Vot)/2
Vf^2= Vo^2 + 2aS
Vf= Vo +at

3. The attempt at a solution

ive tried numerous ways, but i cant seem to get past point of the final velocity at the peak being zero to solve for time in the last equation...and i cant figure out what to do from there
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 18, 2007

### qspeechc

The motion is in two directions; the x- and y-directions. At the peak the speed in the y-direction is zero, but there is still speed in the x-direction. You need to analyse this problem in two-dimensions.

3. Oct 18, 2007

### Yannick

Hi

To prove that the 60° shot will travel 3 times as high as the 30°:

You can use the conservation of energy: Kinetic Engergy at the time of the shot equals the potential energy at the peak.

$$E_{kin}=E_{pot}$$

For $$E_{kin}$$ you have to take the angle into account:
$$E_{kin}=\frac{1}{2}*m*v_{0}^{2}*sin(\alpha)^{2}$$

Solve for h (=max height) and compare the values for 30 and 60 degrees.

I hope this helps...