Recent content by rogers236

I've done that and get: (-2V0y)/(-9.8) = Time-y 1000/V0x = Time-x then to ((-9.8)1000)/V0x = -2V0y ((-9.8)1000)/(1500cos\Theta) = -2(1500)sin\Theta ((-9.8)1000) = -2(1500)sin\Theta(1500cos\Theta) ((-9.8)1000) = (-2)(1500)(1500)sin\Thetacos\Theta...

Thanks for the reply. I checked out that link and then fixed my original post a bit, putting in the equations as I learned them.

Homework Statement A sniper barrel is exactly one meter above the ground and perfectly horizontal. 1000m away is a target one meter above the ground. if the bullet leaves the gun with a muzzle velocity of 1500m/s will it reach the target? (no, next part is my problem) at what angle should the...