# Horizontal projectile

1. Feb 7, 2005

### xelda

A hunter aims directly at a small target (on the same level) 68 m away. If the bullet leaves the gun at a speed of 186 m/s, by how much will it miss the target?

I just need to know what formula I should use to solve this problem?

2. Feb 7, 2005

### Galileo

First calculate the time it takes for the bullet to traverse the given distance.
Then calculate the amount by which the bullet dropped due to gravity.

3. Feb 7, 2005

### xelda

I understand that's how to solve the problem, but I still don't think I'm doing this correctly. To calculate the time, I took 186 m/s divided by 68 m and got 2.735 s. Then I plugged the time into the equation y=yo+vyot-1/2gt^2. So it looked like 0+186m/s(2.735s)-1/2(9.8m/s^2)(2.735s)^2. I ended up with 472 but this number seems too large? Am I doing something wrong here?

Thank you for the quick response. I appreciate your help.

4. Feb 7, 2005

### Galileo

You seem to have confused a few things.
You have to treat the horizontal and vertical components independently.
First, since there is no horizontal acceleration, the horizontal velocity is constant.
To get the time traveled, use d=vt. (d=distance, v=velocity, t=time) (so t=d/v)

Then for the vertical component, you can use $y(t)=y_0+vt-1/2gt^2$, but
think about what v is in this equation. It's the initial component of the velocity in the y-direction.

5. Feb 7, 2005

### xelda

Okay, I used 9.9 m/s^2 for v and my final answer came to 2.93 m. Does this seem more accurate? I guess I was way off. :x

6. Feb 7, 2005

### Galileo

What value did you find for t? And what is the initial v in the vertical direction?

7. Feb 7, 2005

### xelda

For t I got .3656 s. I put 0 in for yo but should v also be 0 if it's the vertical component?

8. Feb 7, 2005

### Galileo

If the target is on the same level as the gun, then isn't the initial velocity horizontal?

9. Feb 7, 2005

### xelda

Oh, okay. Now it's starting to make more sense. Since the initial velocity is horizontal, then it would be 0.