# Trajectory with a Gun

1. Dec 8, 2008

### roboman32

I am having a problem with this question. I have it half solved but I cannot figure out what equation I am supposed to use for the rest.

1. The problem statement, all variables and given/known data
A hunter shoots his rifle at an angle of 27.2 degree above the horizontal. He is holding the rifle so that the muzzle of the rifle is exactly 2 meters above the ground. The muzzle velocity of the rifle is 412 m/s

The answers needed for this question are the following:
How long is the bullet in the air?
What is the bullet's range?
What is the maximum height of the projectile?
How long does it take the bullet to reach its maximum height?

Starting velocity = 412 m/s
Starting height = 2 meters
Range = 14079.67
change in time = 38.4 seconds

2. Relevant equations

y=y0+Y0y$$\Delta$$t+1/2(ay)$$\Delta$$t2
range=v0(cos27.2)$$\Delta$$t

3. The attempt at a solution
I have found the answers to the first two answers which are:
How long is the bullet in the air? 38.4 seconds
What is the bullet's range? 14079.67 meters

but I am not sure what equations I am supposed to use for these questions:
What is the maximum height of the projectile?
How long does it take the bullet to reach its maximum height?

2. Dec 8, 2008

### LowlyPion

Welcome to PF.

You can calculate the time to max height easily enough.

Vo*Sinθ = g*t

t = Vo*Sinθ/g

You should know then that Y = 2 + ½gt²

3. Dec 8, 2008

### roboman32

thank you

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