# Homework Help: Projectile Motion

1. Feb 1, 2010

### black_hole

1. The problem statement, all variables and given/known data

I need help with these 2 problems:

A hunter aims directly at a target (on the same level) 81 m away. If the bullet leaves the gun at a speed of 225 m/s, how far below the target will the bullet hit?

A rifle bullet is fired at an angle of 15.2° below the horizontal with an initial velocity of 154 m/s from the top of a cliff 63.9 m high. How far from the base of the cliff does it strike the level ground below?

2. Relevant equations

3. The attempt at a solution

Vx = 225 m/s, x = 81 m
Vyo = 0 m/s ?
a = -9.8 m/s/s, t = .36 s (from Vx = x/t)

so I used y = Vot + at^2 to find y, but that's not right...

and for the second one

Vx = 148.613 m/s
Vo = -40.377 m/s, a = -9.8 m/s/s, y = -63.9 m
using the same equation, t = 1.221 s
and V = x/t, x= 181.456 m

2. Feb 1, 2010

### Staff: Mentor

In the first one, you're just missing a 1/2 in your equation here:

3. Feb 1, 2010

### black_hole

Wow, how embarrassing. However, in the 2nd one. I now get 201.965 m, which is not right?

4. Feb 1, 2010

5. Feb 1, 2010

### black_hole

well, using the correct equation y = Vo + 1/2at^2 and keeping everything the same
Vx = 148.613 m/s
Vo = -40.377 m/s, a = -9.8 m/s/s, y = -63.9 m
I found t = 1.359 s
and V = x/t, x= 201.965 m...

6. Feb 1, 2010

### Staff: Mentor

The bullet in the 2nd problem is fired downward, so it's initial y velocity is not zero...

7. Feb 1, 2010

### black_hole

I meant y = Vot + 1/2at^2
I'm sorry if I was not clear, but Vyo = -40.377 m/s

8. Feb 1, 2010

### Staff: Mentor

That looks mostly right then, unless there is a small rounding error or something. Is it not the right answer? If not, could you please write out each step again, and I;ll do a more careful check of the math. (I'm bailing for a couple hours though -- will try to check back a bit later)