Posted by **Sarah** on Thursday, June 4, 2009 at 9:06pm.

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?

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**MathMate**, Thursday, June 4, 2009 at 10:00pm
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 -
**Sarah**, Thursday, June 4, 2009 at 10:38pm
what is U? and i dont have time so i need to find it and what do i put in for S to find t? sorry im so confused

- Physics -
**MathMate**, Thursday, June 4, 2009 at 10:44pm
u is the initial velocity = muzzle velocity.

The formula is to find S, the distance the bullet drops vertically.

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**Sarah**, Thursday, June 4, 2009 at 10:55pm
so u would = 750 ft/s, but there is no time so how can you solve for S if t is not available

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**MathMate**, Thursday, June 4, 2009 at 11:01pm
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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