Physics
posted by Sarah .
A markman fires a .22 caliber rifle horizontally at a target the bullet has a muzzle velocity with magnitude 750 ft/s. How much does the bullet drop in flight if the target is (a) 50.0 yd away and (b) 150.0 yd away?

Since he shoots horizontally, the drop starts the moment the bullet leaves the rifle, at the rate of g (acceleration due to gravity).
For (a) and (b), calculate the time t the bullet is in the air using the horizontal speed, then calculate drop using
S=ut+at^2/2
where a=g, and u the muzzle velocity. 
what is U? and i don't have time so i need to find it and what do i put in for S to find t? sorry im so confused

u is the initial velocity = muzzle velocity.
The formula is to find S, the distance the bullet drops vertically. 
so u would = 750 ft/s, but there is no time so how can you solve for S if t is not available

The time is simply the horizontal distance divided by the muzzle speed, since the rifle is aimed horizontally, so cos0=1.