Hi folks, thanks for looking at this. I've tried doing this and revisiting it to look for some fresh thoughts but can't seem to figure out how to use the suvat equations to determine the deviation from the centre of the target.

 

A gun is aimed so that it points directly at the centre of a target 200m away. If the bullet travels at 200m/s how far below the centre of the target will the bullet hit?

My aim is to determine the angle of the trajectory by using suvat for horizontal motion i.e. s = v*cosTHETA*t, but given v, I still have to contend with s, cosTHETA and t. I was thinking of using simultaneous equations with displacement and distance, however, I cannot figure of a way to determine the distance travelled by the projectile. If I use the suvat equation s = ut + 1/2at^2, the t and s are still missing. Can anyone give me a hint if I am on the right track or if not, as to how I should start?

Any assistance is greatly appreciated, thanks! ~

Edited December 23, 201213 yr by cxk216

Well, first of all, you need to make some sort of assumption about the angle of the aim (and from that, bullet velocity), most likely horizontal. That said, you can figure out your s and t values using what you know.

I agree, bullet velocity should be horizontal, however, the effect of gravity leads to a change in distance which the bullet needs to travel. Therefore aiming straight at the bulls eye, the displacement between the barrel of the gun and the centre of the target is 200m, the hypotenuse formed by the displacement from barrel of gun to actual bullet contact with bulls eye will be incrementally bigger. The angle should be small but not negilible. Hence, given this, how can the suvat equations be used given that either the theta or the t is required to solve.

You have v and d, so t is easily determined. You use the components, and gravity only affects the y components, not the x, because the force is in the y direction.

I agree, bullet velocity should be horizontal, however, the effect of gravity leads to a change in distance which the bullet needs to travel. Therefore aiming straight at the bulls eye, the displacement between the barrel of the gun and the centre of the target is 200m, the hypotenuse formed by the displacement from barrel of gun to actual bullet contact with bulls eye will be incrementally bigger. The angle should be small but not negilible. Hence, given this, how can the suvat equations be used given that either the theta or the t is required to solve.

But why should gravity effect the horizontal component of the velocity? I presume we are ignoring air resistance etc.

 

How far away is the board (horizontal component only) - what velocity does the bullet have (horizontal only)? You now know how long the flight time is.

I agree, bullet velocity should be horizontal, however, the effect of gravity leads to a change in distance which the bullet needs to travel. Therefore aiming straight at the bulls eye, the displacement between the barrel of the gun and the centre of the target is 200m, the hypotenuse formed by the displacement from barrel of gun to actual bullet contact with bulls eye will be incrementally bigger. The angle should be small but not negilible. Hence, given this, how can the suvat equations be used given that either the theta or the t is required to solve.

 

But the vertical velocity component that the bullet acquires (from gravity) does not affect the horizontal velocity. The horizontal distance and the horizontal velocity remain constant. Otherwise, we're getting into calculus, as you may have suspected. If the target is dropped at the instance the bullet leaves the barrel, the bullet will strike the bull's eye, honest.

 

We are, of course, neglecting the effects of air resistance, otherwise there's tons of effects to account for. Plus, we're assuming a flat earth for such a short distance (200m), otherwise all bullets technically travels elliptical orbits near a spherical earth, and trust me, you don't want to go there either.