physics problem help please

[Turbo3000]

I just had my test and I kept overthinking the following problem. This isnt for homework or anything, so I don't think it is dishonest. I asked some fellow classmates about it and they were confused as well. I think it is most likely simple and I am overthinking it.

A rifle is fired horizontely at 400 m/s. A target is 200m away. How far below the "bullseye" does the bullet hit?

[Mr incredible]

depends what the bullet drop pattern is, which wil be a factor of aerodynamics, grain weight, and velocity.

You know the velocity, which doesn't look very fast BTW, now you need to know the grain weight and aerodynamics (hollow point, target, hunting)

[Turbo3000]

That is the only given information. It should be solvable with just that info, that is what the prof. said.

[Mr incredible]

http://www.worsleyschool.net/science...rajectory.html

this should help but I have to make dinner so aint got the time

[Scrumhalf]

Easy.

The key to these problems is to remember that the movement of the bullet along any arbitrary trajectory can be decomposed into movement in two orthogonal directions, i.e. horizontal movement and vertical movement.

1st Step:
Calculate how much time it takes the bullet to move 200m horizontally. Since it is moving at 400m/sec, it takes 0.5sec.

2nd Step:
How much does the bullet drop vertically in 0.5 sec? Since horizontal movement and vertical movement can be treated independent of each other, you can do this separately.

The intial velocity of the bullet in the vertical direction is zero. Remember that:

s = u*t + 0.5*a*t^2

where u = initial velocity = zero
a = acceleration due to gravity
t = time
and ^2 represents the square

So in this case,

s = 0.5 * a* t^2

= 0.5*9.8*0.5^2
= 0.5*9.8*0.25
= 1.22m

So, the bullet will drop 1.22m in 0.5 sec.

[Mr incredible]

LOL, so whats the answer then Scrumhalf smarty pants?

BTW you seen we got Farrell at Fly half against SA, crazy but interesting

[ByRoN]

easy my ass!!!! that shit made my head hurt!!!

Originally Posted by Scrumhalf

Easy.

The key to these problems is to remember that the movement of the bullet along any arbitrary trajectory can be decomposed into movement in two orthogonal directions, i.e. horizontal movement and vertical movement.

1st Step:
Calculate how much time it takes the bullet to move 200m horizontally. Since it is moving at 400m/sec, it takes 0.5sec.

2nd Step:
How much does the bullet drop vertically in 0.5 sec? Since horizontal movement and vertical movement can be treated independent of each other, you can do this separately.

The intial velocity of the bullet in the vertical direction is zero. Remember that:

s = u*t + 0.5*a*t^2

where u = initial velocity = zero
a = acceleration due to gravity
t = time
and ^2 represents the square

So in this case,

s = 0.5 * a* t^2

= 0.5*9.8*0.5^2
= 0.5*9.8*0.25
= 1.22m

So, the bullet will drop 1.22m in 0.5 sec.

[Scrumhalf]

Originally Posted by Mr incredible

LOL, so whats the answer then Scrumhalf smarty pants?

BTW you seen we got Farrell at Fly half against SA, crazy but interesting

The bullet will hit 1.22m below the bullseye - I said that in my earlier post .. lol..

[Lmg2701]

Yea, I don't think the problem is attempting to be difficult. It's basically just making sure you take in account the gravity pushing the bullet down. Once you find the right equation you will just have to plug in the 9.81 m/s^2 and the rest of the info is given.

[Turbo3000]

Thanks everyone, I actually got an answer similar to that. I ended up finding the degree of the angle created by dropping bullet, I then used that in some other equation to get the range, and took the range it traveled and subtracted it from the range the target was located. So, maybe I will get credit for it, but I did it the wrong way.

[magdalena]

You never know, since you said that you got a similar numbers...sometimes there are many ways to come up with the right answer.

[rado]

Originally Posted by Scrumhalf

Easy.

The key to these problems is to remember that the movement of the bullet along any arbitrary trajectory can be decomposed into movement in two orthogonal directions, i.e. horizontal movement and vertical movement.

1st Step:
Calculate how much time it takes the bullet to move 200m horizontally. Since it is moving at 400m/sec, it takes 0.5sec.

2nd Step:
How much does the bullet drop vertically in 0.5 sec? Since horizontal movement and vertical movement can be treated independent of each other, you can do this separately.

The intial velocity of the bullet in the vertical direction is zero. Remember that:

s = u*t + 0.5*a*t^2

where u = initial velocity = zero
a = acceleration due to gravity
t = time
and ^2 represents the square

So in this case,

s = 0.5 * a* t^2

= 0.5*9.8*0.5^2
= 0.5*9.8*0.25
= 1.22m

So, the bullet will drop 1.22m in 0.5 sec.

:bravonew:

[THE BOUNCER]

Originally Posted by Scrumhalf

Easy.

So, the bullet will drop 1.22m in 0.5 sec.

agreed. the problem is so easy i didnt even feel like responding to this none sense thread.

[coffee-guy]

I just aim a little higher as I get farther away from the target - I would suggest aiming about 1.22mm higher than you typically would in this case ....