A Rifle shoots bullets @ 420m/s is to be aimed @ 42.0m away.?

Question

If the center of the target is level with the rifle, how high (in cm) above the target must the rifle barrel be pointed so that the bullet hits dead center?
physics problem.
I try many times with no luck, need help!!!!!

Thank you guys!! now I understand better!!!

Answer 1

I have a different method which I think answers the question as it is asked.
The rifle is at the same level as the target, but the barrel is pointed upwards. That means the horizontal velocity will be ever-so-slightly less than 420m/s.
Let the point that the rifle is pointed at be 'x' m above the target.
Let the angle that the gun barrel makes with the horizontal be 'θ'.
The vertical velocity of the bullet is 420sinθ
The horizontal velocity of the bullet is 420cosθ.
The time taken for the bullet to reach the target is therefore t=42/420cosθ.
In half the same time, the vertical velocity of the bullet must have been reduced to zero by gravity.
0.5t=420sinθ/9.8 or t=840sinθ/9.8
Therefore
42/420cosθ=840sinθ/9.8
sinθcosθ=1.167E-3

sinθcosθ=sin2θ/2 (a "trig identity")

sin2θ/2=1.167E-3
θ=0.0669 degrees

Using tangent to find x

tan0.0669 degrees = x/42
x=0.049m or 4.9cm

The answer is the same as gp4rts because the speed of the bullet is so great. But if you were to try it with a slower projectile you would see the two methods diverge somewhat, as the rifle would have to be pointed at a greater angle so the horizontal component would decrease significantly.

Answer 2

The bullet will fall under the influence of gravity, accelerating at g. In time T, it will fall a distance s = 0.5*g*T^2. Find T by determining how long it will take the bullet to travel the horizontal distance to the target. The bullet velocity is 420 m/sec, the target is 42 m away, so it will take 42 / 420 = 0.1 seconds. That is T.

It will fall 0.5*9.8*0.1^2 = 0.049 m. The rifle must be that much higher than the target.