Posted by Marcus Jasp on .
A gun shoots bullets that leave the muzzle at 250 m/s. If a bullet is tohit a target 160.8 m away at the level of the muzzle, the gun must be aimedat a point above the target. (Neglect air resistance.)
How far above the target is this point if the angle the gun makes is greater than 45°?
These are the tips they give me:
Use the formula for the range of a projectile.
I spent a whole day on this problem homies...HELP!!!
The range of the projectile is given by d = Vo^2(sin(2µ)/g where Vo = the initial muzzle velocity, g = the acceleration due to gravity and µ = the angle of the initial velocity vector aboc the horizontal. You have 3 of the pieces of information required. Substitute and solve for µ.