- #1

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## Homework Statement

A monkey is standing on the top of a cliff 50 metres high and drops a coconut. His friend is standing at the base of the cliff and throws a coconut upward with an initial velocity of 15m/s. At what distance and at what time will the two coconuts collide?

## Homework Equations

v

^{2}= u

^{2}+ 2as

## The Attempt at a Solution

__Monkey throwing down:__

v

^{2}= u

^{2}+ 2as

v

^{2}= 2(9.81m/s

^{2})(50m)

v

^{2}= 981.m/s

v = 31.m

a = (v

_{2}- v

_{1}) / t

a = (31.m/s - 0m/s) / t

a = 31.m/s / t

t = 31.m/s / 9.81m/s

^{2}

t = 3.1s

__Monkey throwing up:__

v

^{2}= u

^{2}+ 2as

0m/s = (15m/s)

^{2}+ 2(-9.81m/s

^{2})s

0m/s = 225m/s - 19.6m/s

^{2}s

-225m/s = -19.6m/s

^{2}s

s = 11.5m

a = (v

_{2}- v

_{1}) / t

-9.81m/s

^{2}= (0m/s - 15m/s) / t

-9.81m/s

^{2}= -15m/s / t

t = -15m/s / -9.81m/s

^{2}

t = 1.5s

I have the initial and final velocities, the distances, the acceleration, and the times for both coconuts. I'm not sure how to find when the balls actually meet and where they meet. At first, I thought you were supposed to subtract the two times but then I realized that doesn't actually give you the time when the coconuts collided. Thank you in advance for your help.