you have a rifle and you have a target, with the bullseye at the same height as the barrel, that is 50 m away, calculate the angle at which you need to fire the gun to get a bullseye.

I'm pretty sure you need the velocity of the projectile to work that out. Unless there's a universally accepted rifle projectile speed?

 

Either way I can't solve it for you, but just in case you do need more information (and now I'm curious as to how to figure it out lol.)

you would need to raise the rifle a little higher, due to wind resistance. However, you need velocity to see the power of the bullet as it leaves the gun.

oh! i'm sorry! velocity is 438 m/s

s = v₀t + 1/2 at²

 

Use this for both the horizontal and vertical, find the initial vertical velocity needed, and then add the vectors.

so it`ll hit the target in roughly 1/9`th of a second.

so if you have the perfect gun (and it doesn`t kick up or down on firing, and the barrel remains stationary) then using the rate of fall and acceleration towards gravity, how far will a bullet drop when "let go" in 1/9`th of a second?

find that, and there`s you`re answer

 

fact is it won`t be far, I`de guess at most you`de need to aim roughly an inch above the bullseye (assuming the perfect gun).

s = v₀t + 1/2 at²

 

Use this for both the horizontal and vertical' date=' find the initial vertical velocity needed, and then add the vectors.[/quote']

it doesn't give you the horizontal and the vertical componants just an initial velocity and how far it has to go!

nevermind i got it <stupid satan question!>

 

thanks anyway!

The bullet is long gone and hit the target before the gun even moves. So it doesnt matter if your gun kicks or not, that doesnt throw off a shot.

 

I wonder if a bullet drops at the speed of earths gravity when its traveling at 438m/s through the air. I would Imagine the air would somehow support it and slow the drop by a slight amount.

The bullet is long gone and hit the target before the gun even moves. So it doesnt matter if your gun kicks or not' date=' that doesnt throw off a shot.

 

I wonder if a bullet drops at the speed of earths gravity when its traveling at 438m/s through the air. I would Imagine the air would somehow support it and slow the drop by a slight amount.[/quote']

 

most questions like these negate air resistance

 

*unless its an actual question he wanted to know opposed to some question he had to do for a physics class

The bullet is long gone and hit the target before the gun even moves. So it doesnt matter if your gun kicks or not' date=' that doesnt throw off a shot.

 

I wonder if a bullet drops at the speed of earths gravity when its traveling at 438m/s through the air. I would Imagine the air would somehow support it and slow the drop by a slight amount.[/quote']

 

You Are kidding me right!

 

and as for the second part any bullet I`ve ever seen and used is symetrical along its center and so any upforce from air will be met with equal down force, and if the barrel`s rifled it wouldn`t make a whole hell of alot of difference if it were a fraction off center anyway

You will also need to know the ballistic coefficient of the bullet in order to ascertain the rate at which it loses velocity due to air resistance.

 

It will not be moving as fast when it gets to the target as it was initially.

If you shoot a bullet from a rifle which is 1m above the ground and drop a bullet from 1m above the ground both bullets will hit the ground at the same time.

 

Reason, the fact that one bullet is moving at a few hundred meters per second does NOT change the speed of gravity. All objects will accelerate downwards (due to gravity) at 9.8m/s^2 regardless of their horizontal velocity.

If you shoot a bullet from a rifle which is 1m above the ground and drop a bullet from 1m above the ground both bullets will hit the ground at the same time.

 

Reason' date=' the fact that one bullet is moving at a few hundred meters per second does NOT change the speed of gravity. All objects will accelerate downwards (due to gravity) at 9.8m/s^2 regardless of their horizontal velocity.[/quote']

 

We'll call this the "wingless bullet" assumption!

the funny thing is.... is that no one said we had to be on earth to hit the target.... we could be in space and these calculations would'nt quite work well...... uhhh at least i think so....

We'll call this the "wingless bullet" assumption!

do excess velocities change the equation. like lets say the bullet travels 5 miles a second. the earth curves at 8 in at a mile away. so if you travel a mile in any given direction you'll be 8 inches lower than at your original point.

Right?

do excess velocities change the equation. like lets say the bullet travels 5 miles a second. the earth curves at 8 in at a mile away. so if you travel a mile in any given direction you'll be 8 inches lower than at your original point.

Right?

 

The assumption is that you are shooting parralel to the ground or at a tangent to the "perfect surface" so ignoring air resistance if you shoot faster than orbit speed at that height your shot will rise, not fall.

 

If you shoot exactly at orbit speed, don't forget to duck!

The assumption is that you are shooting parralel to the ground or at a tangent to the "perfect surface" so ignoring air resistance if you shoot faster than orbit speed at that height your shot will rise' date=' not fall.

 

If you shoot exactly at orbit speed, don't forget to duck![/quote']

 

 

you're right, the shot will rise because of the tangent ( not quite sure what a tangent is) but no one said that we had to assume that we were on earth.

It is generally assumed that problems such as these do not take in "earthly considerations" such as air resistance. If air resistance were taken into consideration, it would further complicate the problem.