Hi, I need help with this physics problem, I'm not quite getting it. Do you?

Question

A bullet was fired at a target 50 m away with a muzzle velocity of 300 m/s. If the barrel of the gun was horizontal, how far in cm below the level of the gun did the bullet
strike the target?

Oh yes, It would be nice if the formula used for solving this problem was shown in the answer.

Answer 1

the way to work these problems is this. first, split the motion into horizontal and verticle problems. Then think about the fact that time is the same for both halves.

in this case..

******horizontal motion*****

v = d/t
t = d / v

where t= time, d = distance, v = velocity

t = 50 m / (300 m/s) = 0.167 seconds

***** verticle motion ****

from here
http://en.wikipedia.org/wiki/Equations_of_motion

x = Vot + 1/2 a t ^2

Vo = 0
a = 9.8 m/s^2
t = .167 seconds (from horizontal motion)

x = 0 + 1/2 *9.8 m/s^2 * (.167 sec)^2 = 0.137 meters

= 0.137 m x (100 cm/m) = 13.7 cm

Answer 2

Since you asked for the answer in CM :

The bullet strikes the target 13.6 cm below the level of the gun.

At 300 m/s, the bullet will take 1/6 s to reach the horizontal position of the target, 50 m away. (Since t = d/v = 50/300 s
= 1/6 s.)

During that time it will fall 1/2 g t^2 = 1/2 (9.80) (1/6)^2 m

= 0.1361... m = 13.61...cm.

So the bullet strikes the target 13.6 cm below the level of the gun, to the 3 significant figures justifed by the input data. (One could actually argue that only 14 cm was justified, as "50 m" is only given to TWO sig. figs.)

Live long and prosper.

P.S. Note that I made a small mistake in expressing the numerics of my initial answer, which the responder immediately below, 'Two v's...' ("a devout Christian and I strongly believe God's Word"), then pasted into her own answer. She also pasted in my very own original words, both leading up to those numerics and those used immediately thereafter in describing the horizontal motion. Hmmm !! It's interesting to note that God's devout follower is a plagiarist.

It is in fact OUTRAGEOUS PLAGIARISM!

Answer 3

It will take 1/6 s to reach the horizontal position of the target.

During that time it will fall 1/2 (9.80) (1/6)2 m =

0.1361