phyisics
posted by Carolina on .
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!!

Go to the bottom coconut
20=Vi*t1/2 g t^2
Now go to the top coconut
20=401/2 g t^2
solve for t in the second equation, put it in the first, solve for vi 
waht is g?

and what do you mean by top and bottom?
thanks 
top of clift, bottom of cliff
g is 9.8m/s^2, the acceleration of gravity. 
why didyou use
these equations
Go to the bottom coconut
20=Vi*t1/2 g t^2
Now go to the top coconut
20=401/2 g t^2 
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 
If objects acted simultaneously their relationship is time. The time is obtain because the time of collision between the coconut is the same.