A projectile is fired with an initial muzzle speed 190 m/s at an angle 60 from a position 5 meters above the ground level.

Find the horizontal displacement from the firing position to the point of impact.

I know that y=vsin(60)t-.5gt^2 and x=vcos(60)t. but I keep getting wrong answer.

3 answers

so, what do you get?
Try plugging in you numbers here and see whether it is the same. The range is

R = v^2/g sin2θ
But does the range equation still work if the projectile starts 5 feet off the ground.