# Physics

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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?

• Physics -

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.

• Physics -

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

• Physics -

u is the initial velocity = muzzle velocity.
The formula is to find S, the distance the bullet drops vertically.

• Physics -

so u would = 750 ft/s, but there is no time so how can you solve for S if t is not available

• Physics -

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

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