# Oh, My Brain!

1. Sep 14, 2004

### physicsss

A shooter aims directly at a target on the same level 200 m away.

a) If the bullet leaves the gun at a speed of 250 m/s, by how much will it miss the target?

b) At what angle should the gun be aimed so the target will be hit?

2. Sep 14, 2004

### recon

The bullet will travel horizontally at 250 m/s, thus making the nearest encounter with the target at 200/250 = 0.8 s. In this time, how much will the bullet have fallen from its original position? Remember that the acceleration due to gravity is 9.8 m/s2. The answer to (b) should immediately follow.

3. Sep 14, 2004

### physicsss

For part a, I used the equation y2=y1+v1*t-1/2*g*t^2
since y1=0 and v1=0 in the y-direction, the equation is left with y=1/2*g*t^2
y=-1/2*(-9.8)*0.8^2=3.14 (is my sig figs correct?)

For b, I used v^2=v1^2-2gy, since at halfway distance is 100m and v=0 in the y direction, equation becomes

0=(sin(x)*250)^2+2*9.80*100

x=10.2
Is this and the sig figs right?

4. Sep 15, 2004

### Chronos

Last edited: Sep 15, 2004
5. Sep 15, 2004

### physicsss

Last edited: Sep 15, 2004
6. Sep 15, 2004

### Tide

Actually, the nearest encounter will occur somewhat earlier than that!

But you're really interested in the time it takes for the horizontal distance travelled by the bullet to equal the original distance to the target.

7. Sep 15, 2004

### recon

Yeah, that's what I meant. I just didn't know how to put it in words. So I opted to put it the way I did, without realising that it was wrong.