Question

A monkey in a perch 20m high in a tree drops coconut above the head of a zoo keeper as he runs with a speed of 1.5m/s beneath the tree how far behind him in meters does the coconut hit the ground? please help me

Answer 1

wat is the speed at which the coconut is falling?

Answer 2

\[20-(9.8/2)t^2=0\]
\[\sqrt{20/4.9}\]
\[\sqrt{20/4.9}*1.5=3.03m \]

Answer 3

This may help you as well...for constant acceleration (ignoring drag etc):
\[distance = 1/2 g t ^2\]
So you can simply plug in values to get this equation:
\[20m = 1/2(9.8 m/s^2)t^2\]Isolate t
\[(20m * 2)/9.8m/s^2=t^2\]Take the square of both sides to solve for t which is how long it will take for the coconut to hit the ground (2.02 seconds). From there, you simply figure out where the monkey will be in 2.02 seconds:
\[2.02s *1.5m/s =3.03m\]