Question

A monkey sitting on a tree branch 35 meters above the ground drops a coconut. at the instant of the drop a man running at a uniform speed of 1.5 meters per second crosses the three. how much earlier should the monkey have dropped the coconut so as to hit the man

Answer 1

Explanation:

u=0

g=10m/s²

s=35m

35=(1/2)*10*t²

therefore, t=√7sec

so the man should have thrown the coconut√7 secs earlier

for solving this , I think height of the man is required...then s=(35-h)m and not 35m ... then the coconut will hit the man in his head

But if we ignore the height of the man,

before √7secs the man will be √7*1.5m behind the tree =1.5√7m

Hope it helps you