Question

An unfortunate 18 kg monkey falls from a 40 m tall tree. What is the monkeys final velocity just befor he impacts the ground.? a 7063.2 m/s b 784 m/s c 28 m/s d 720 m/s

Answer 1

The correct answer is c) 28 m/s.
Let's find the step-by-step solution. The motion of the monkey is an uniformly accelerated motion, with acceleration equal to $g=9.81 m/s^2$. The initial velocity of the monkey is zero, while the distance covered is S=40 m. Therefore, we can use the following relationship to find vf, the final velocity of the monkey:
$2aS=v_f^2-v_i^2=v_f^2$
from which
$v_f= \sqrt{2aS}= \sqrt{2\cdot 9.81 m/s^2 \cdot 40 m}=28 m/s$