Question

A person on ground takes aim at a monkey sitting on a tree which is at some distance, and fires a bullet. Seeing the smoke, the monkey begins to fall freely, then the bullet will

• A. always hit the monkey.
• B. go above the monkey.
• C. go below the monkey.
• D. hit the monkey if the initial velocity of the bullet is more than a certain velocity.

Answer 1

The correct option is D hit the monkey if the initial velocity of the bullet is more than a certain velocity.

The monkey starts a vertical distance h above the hunter, and the horizontal distance between the monkey and hunter is d. The origin is the hunter.

The horizontal position of the bullet is:

$x_b=v_{ox}t$

When $x_b=d,t=d/v_{ox}$

The vertical position of the bullet is:

$y_b=v_{oy}t-½g{t}^2$

Plugging in $t=d/v_{ox}$ for the first $t$ gives:

$y_b=d\frac{v_{oy}}{v_{ox}}-½g{t}^2$

Compare this to the vertical position of the monkey:

$y_m=h-½g{t}^2$

If the hunter aims the gun at the monkey to begin with:

$\tan \left(q\right)=h/d=v_{oy}/v_{ox}$

So, $\frac{d{v}_{oy}}{v_{ox}}=h$

In this case, when the bullet has traveled a distance d horizontally, its vertical position will be:

$y_b=h-½g{t}^2$

That exactly matches the monkey's position, so the bullet will hit the monkey when the hunter aims the rifle at the monkey.