Question

Thought Experiment: A monkey escapes from a zoo and climbs a tree. After failing to entice the monkey down, a zookeeper fires a tranquilizer dart directly at the monkey. The monkey lets go at the same instant the dart leaves the gun barrel, intending to land on the ground and escape. Explain how the dart will always hit the monkey regardless of the earth’s velocity (provided that the dart gets to the monkey before it hits the ground). Draw a picture to assist your explanation.

Answer 1

Explanation:

When bullet is shot towards the monkey then let say the distance of monkey from the bullet is "d"

so we can find the time to reach the bullet to the monkey

$t = \frac{d}{vcos\theta}$

Now similarly we can find the vertical displacement of the bullet in the same time

$\Delta y = vsin\theta t - \frac{1}{2}gt^2$

$\Delta y = v sin\theta (\frac{d}{vcos\theta}) - \frac{1}{2}gt^2$

so it is given as

$\Delta y = d tan\theta - \frac{1}{2}gt^2$

here if the monkey is initially at height H above the ground at given angle then we can say

$H = dtan\theta$

so we can say that

$\Delta y = H - \frac{1}{2}gt^2$

So if at the same time monkey will fall down then the height of monkey from ground after time "t" is given as

$\Delta y = H - \frac{1}{2}gt^2$

so here bullet will hit the monkey as both monkey and bullet are at same position.