# Thread: Help with word problem involving an angle of elevation

1. ## Help with word problem involving an angle of elevation

I have been trying to solve this problem , but so far, I don't know how to use the information I have. In other words , I don't know how to set up this problem!

If muzzle velocity of a rifle is 300 feet per second , at what angle of elevation( in radians) should it be aimed for the bullet to hit a target 2500 feet away?

any help would be good because I'm a little lost. Thanks in advance

2. Originally Posted by skorpiox
If muzzle velocity of a rifle is 300 feet per second , at what angle of elevation( in radians) should it be aimed for the bullet to hit a target 2500 feet away?

any help would be good because I'm a little lost. Thanks in advance
Hokay.
Draw a backwards L. Connect the two lines with a third line. You should now have a right-angled triangle.

The SPEED of the rifle is the diagonal line. It is 300.
The DISTANCE to the target is the horizontal line. It is 2500.

Imagine the gun is on the bottom-left corner of the triangle. The angle of elevation is this corner of the triangle. Angles of elevation are always measured UP from the horizontal.

Using simple trigonometry:

cos(theta)=opposite/hypotonuse
cos(theta)=2500/300
theta=inversecos(2500/300)

:]

Edited: clearer explanation of the triangle

### pdffind the angle of elevation when muzzle speed n target is shown

Click on a term to search for related topics.