a rifle shoots a bullet 460 m/s aimed at a target, if the center of the target is leveled with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

Question

anonymous · Answer

You first have to figure out how long it will take for a bullet to get to the target. Then you have to check how far it will fall in that time and that is your answer. Formulaes to help you along is 
$$v = \frac{s}{t}$$
and 
$$s= v_0 t + \frac{1}{2}g*t^2$$
Try to work with this and if you need more help then ask:)

anonymous · Answer

Plitter's answer above is not quite right and also, there is not enough info to possibly determine an answer. We *need* to know how far away the target is. The reason that Plitter's answer is not correct is that the time to reach the target (assuming a distance is known) actually depends on the answer itself: As soon as the rifle is pointed above the target, it affects the horizontal component of velocity: the higher it is aimed, the longer it would take to reach the target. If the distance to the target is provided, then you could use the range formula:

$$R = v^2 \sin (2\Theta) /g$$  in order to find the launch angle theta. With that known, you could use trig to find how much higher above the target it needs to be aimed at.

anonymous · Answer

I might have misinterpreted the question, but I assumed that the vertical speed of the bullet would always be 0. If that was the question then it would just be to figure out how far the bullet will drop in the period of time it would take the bullet to get to the target, otherwise you are of course right, and your interpretation of the question is also more interesting :)