Need Help with Physics Motion calculation problem?

Oh, man, this is too easy and everyone's trying to do it the HARD way!

Both coconuts are in free-fall.  Their relative velocity DOES NOT CHANGE until they strike something (the ground or each other).  Ergo, the two will collide at the earlier of:

1.  The initial distance over the relative speed (50 m / 15 m/sec = 3.33 sec), AND

2.  The time of impact of the upper coconut on the ground.  t = √(2d/g) = √10.2 = 3.2 sec.

So the two will collide when the coconut dropped from the cliff falls onto the one thrown upward which has already reached the ground, and the distance from the ground will be 0.

For monkeys coconut downwards

h-is the height to meeting point from the base

S= Ut + 0.5x g x t^2

50-h= 0.5x g x t^2-----1

to friends coconut downwards

S= Ut + 0.5x g x t^2

h= (-15)xt+ 0.5x g x t^2---2

solve 1 and 2 ull get the answer

Assuming no air friction, top above floor = 4.9*(t^2) + v0t + ok the place t = time considering launch, v0 = preliminary vertical speed, and ok = preliminary top above floor. 4.9 is a pair of million/2 of the gravitational acceleration of earth (approximately 9.8 m/s^2) ok = 0.1651 m, so h = 4.9*(t^2) +v0t + 0.1651 m. while 2.8 seconds elapse after launch, h = 0 m. 4.9*(2.8^2) + v0*2.8 = 0.1651 m

y the monkeys wants to do that. weird.