Question

At the time t = 0, the boy throws a coconut upward (assume the coconut is directly below the monkey) at a speed v0. At the same instant, the monkey releases his grip, falling downward to catch the coconut. Assume the initial speed of the monkey is 0, and the cliff is high enough so that the monkey is able to catch the coconut before hitting the ground. The time taken for the monkey to reach the coconut is

Answer 1

The time taken for the monkey to reach the coconut, t is = H/v₀

Step-by-step explanation:

Let the coordinate of the boy's hand be y = 0. The height of the tree above the boy's hand be H.

Coordinate of where the monkey meets coconut = y

Using the equations of motion,

For the monkey, initial velocity = 0m/s, time to reach coconut = t secs and the height at which coconut is reached = H-y

For the coconut, g = -10 m/s², initial velocity = v₀, time to reach monkey = t secs and height at which coconut meets monkey = y

For monkey, H - y = ut + 0.5gt², but u = 0,

H - y = 0.5gt²..... eqn 1

For coconut, y = v₀t - 0.5gt² ....... eqn 2

Substituting for y in eqn 1

H - y = H - (v₀t - 0.5gt²) = 0.5gt²

At the point where the monkey meets coconut, t=t

H - v₀t + 0.5gt² = 0.5gt²

v₀t = H

t = H/v₀

Solved!