Monkey up a tree

A hunter is trying to kill a monkey by using a bow and arrow. His arrow has an initial velocity of v in a direction of his choice, and the monkey is up a tree at some point (a,b) relative to the hunter's location (0,0), where a and b are both > 0. At the very instant that the arrow is launched the monkey lets go of the branch and tries to avoid the arrow by dropping to the undergrowth and then running away.

How should the hunter aim the arrow?

Originally posted by talzamir
A hunter is trying to kill a monkey by using a bow and arrow. His arrow has an initial velocity of v in a direction of his choice, and the monkey is up a tree at some point (a,b) relative to the hunter's location (0,0), where a and b are both > 0. At the very instant that the arrow is launched the monkey lets go of the branch and tries to avoid the arrow by dropping to the undergrowth and then running away.

How should the hunter aim the arrow?

The hunter should make sure to stand (almost) directly below the monkey, and aim straight up. That way, the monkey can drop all he likes. In mathematical terms, the hunter should ensure that |b| < size of monkey's backside/2.

You may call that cheating the problem, and you'd be right, of course. But if you watch documentaries, you'll see that this is often the method employed by real rainforest hunters with blowpipes: don't fire sideways, fire up. It reduces the margin of error.

Richard

In love and war and monkey hunting, all methods are fair? 🙂 Yes, that would certainly work.

Originally posted by talzamir
A hunter is trying to kill a monkey by using a bow and arrow. His arrow has an initial velocity of v in a direction of his choice, and the monkey is up a tree at some point (a,b) relative to the hunter's location (0,0), where a and b are both > 0. At the very instant that the arrow is launched the monkey lets go of the branch and tries to avoid the arrow by dropping to the undergrowth and then running away.

How should the hunter aim the arrow?

such that the angle 'A' is equal to the arctangent(b/a)....I think. So, right at the monkey.

Originally posted by joe shmo
such that the angle 'A' is equal to the arctangent(b/a)....I think. So, right at the monkey.

Assuming, of course, that air resistance is zero... which it isn't.

Richard

Originally posted by Shallow Blue
Assuming, of course, that air resistance is zero... which it isn't.

Richard

Well...be my guest Richard, or do you prefer Dick?

Originally posted by joe shmo
Well...be my guest Richard, or do you prefer Dick?

Don't be such a thirteen-year-old.

Richard

Originally posted by Shallow Blue
Don't be such a thirteen-year-old.

Richard

If you weren't such a pedant, I wouldn't have to be.

Right at the monkey is the answer I sought. It made for a nice demo at physics class, with a playing card and a small crossbow, with a laser gate just before the crossbow so that the card was dropped at the very moment that the bolt was fired. The card dropped about two feet before it was skewered to the plank behind it. Obviously there are things we can consider to refine this further, such as air resistance, the non-uniformity of the gravitational field, the rate of acceleration of the bolt as it is launched, etc, but currently I'm not in mood to go there. So right at the monkey it is.