Question

an orangutan throws a coconut vertically upward at the foot of a clif 40 m high while his mate simulaneously drops another coconut from the top of the clif. the two coconuts collide at an altitude of 20 m what is the initial velocity of the coconut that was thrown upward?


i know the equations

x= position = .5a t^2 + v (velocity) initial t + x initial (pos)

v= at+v initial
a=a

please help thanks!!

Answer 1

Go to the bottom coconut
20=Vi*t-1/2 g t^2
Now go to the top coconut
20=40-1/2 g t^2
solve for t in the second equation, put it in the first, solve for vi

Answer 2

waht is g?

Answer 3

and what do you mean by top and bottom?

thanks

Answer 4

top of clift, bottom of cliff
g is -9.8m/s^2, the acceleration of gravity.

Answer 5

why didyou use

these equations

Go to the bottom coconut
20=Vi*t-1/2 g t^2
Now go to the top coconut
20=40-1/2 g t^2

Answer 6

You were given distance, those are the distance equation.

finalposition=initial position+vi*t+ 1/2 a*t^2
memorize that. g= is negative(downward), so make the last term - 1/2 9.8 t^2, some just write - 4.9t^2

Answer 7

If objects acted simultaneously their relationship is time. The time is obtain because the time of collision between the coconut is the same.

Answer 8

A coconut is hanging on a tree at height of 15m from the ground a body launches as projectile vertically upward with velocity 20m/sat what time projectile will pass the coconut