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## Homework Statement

A rifle, in which the initial velocity of the bullet is 460m/s, is aimed at a small target 800 metres away at the same height as the rifle. At what angle above the ground must the rifle be pointed so that it hits the target?

**

x = horizontal distance

t = time

V

_{x}= horizontal velocity

g = gravitational acceleration

y = vertical distance

V

_{yi}= initial vertical velocity

V

_{yf}= final vertical velocity

**

so far I know:

x is 800m (given)

g = -9.81m/s

^{2}, since the bullet is being pulled down by gravity

y = 0, because the bullet does not travel any vertical distance because it lands at the same height that it is shot at.

http://img144.imageshack.us/img144/1736/physicsproblem2bs1.jpg [Broken]

## Homework Equations

vertical:

y = V

_{yi}t + .5gt

^{2}

y = V

_{yf}t - .5gt

^{2}

g = (V

_{yf}- V

_{yi})/t

V

_{yf}

^{2}= V

_{yi}

^{2}+ 2gy

horizontal:

V

_{x}= xt

## The Attempt at a Solution

I'm almost clueless on how to approach this one. I can't even see how there is enough info given to be able to solve it. all I have so far is a diagram

http://img464.imageshack.us/img464/4261/physicsproblemvf7.jpg [Broken]

Thanks in advance!

Alex

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