You are standing a far ways away from the base of a tree from which a monkey is hanging on a branch exactly 5 meters above the ground.

This monkey has psychic relaxation syndrome, which means that as soon as the gun goes off, he will let go of the tree branch without realizing it and fall down.

The problem with this is the monkey will fall, and not be where it was before you shot the gun, so it will be in another spot when the bullet is next to the tree. How far below the spot where the monkey's tail is before the shot is fired should you aim so the monkey's tail will be hit regardless of its "letting go" of the tree branch?

(Assume there is zero air resistance.)

Everyone's right about aiming it right at the monkey's tail and neglecting air resistance, it'll hit it there. But there's the proof.

For constant acceleration, the position can be given by x-x0=v0*t+0.5*a*t^2. The problem doesn't state that there's anything causing acceleration/force in the x direction. The only acceleration would be gravity. Let the monkey be at L away from you and hang up H on the tree.

The monkey's position with time is (L, H-0.5*g*t^2). The position of the bullet will be (v0*cos(z)*t,v0*sin(z)*t-0.5*g*t^2). v0 is the speed of the bullet and z is the angle at which we need to determine.

For you to hit the monkey, v0*cos(z)*t=L and v0*sin(z)*t-0.5*g*t^2=H-0.5*g*t^2. This means v0*cos(z)*t=L and v0*sin(z)*t=H --> tan(z)=H/L. Therefore, you gotta aim right for the tail!

Posted by np_rt on 2004-05-30 09:04:00