Posted by **Carolina** on Monday, October 12, 2009 at 5:04pm.

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!!

- phyisics -
**bobpursley**, Monday, October 12, 2009 at 5:08pm
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

- phyisics -
**Carolina**, Monday, October 12, 2009 at 5:18pm
waht is g?

- phyisics -
**Carolina**, Monday, October 12, 2009 at 5:19pm
and what do you mean by top and bottom?

thanks

- phyisics -
**bobpursley**, Monday, October 12, 2009 at 5:31pm
top of clift, bottom of cliff

g is -9.8m/s^2, the acceleration of gravity.

- phyisics -
**Carolina**, Monday, October 12, 2009 at 5:42pm
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

- phyisics -
**bobpursley**, Monday, October 12, 2009 at 5:48pm
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

- phyisics -
**Heisenberg**, Sunday, February 14, 2016 at 11:41am
If objects acted simultaneously their relationship is time. The time is obtain because the time of collision between the coconut is the same.

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