Asked by Carolina

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!!

Answers

Answered by bobpursley
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
Answered by Carolina
waht is g?
Answered by Carolina
and what do you mean by top and bottom?

thanks
Answered by bobpursley
top of clift, bottom of cliff
g is -9.8m/s^2, the acceleration of gravity.
Answered by Carolina
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
Answered by bobpursley
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
Answered by Heisenberg
If objects acted simultaneously their relationship is time. The time is obtain because the time of collision between the coconut is the same.
Answered by Rohan
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
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions