vertical motion?

Question

An orangutan throws a coconut vertically upward at the foot of a cliff 40 m high while his mate simultaneously drops another coconut from the top of the cliff. the 2 coconuts collide at an altitude of 20 m wat was the initial velocity of the coconut that was thrown upward.?

Answer 1

Find two equations describing the motion of each coconut.

First, the coconut which was simply dropped has no initial velocity, so its equation looks like,
d = 1/2at^2
where d is the distance traveled, a is the acceleration experiences, and t is the time

Second, the coconut which is thrown upward has an initial velocity...call it v, its equation looks like,
d = v*t + 1/2at^2

The acceleration each coconut experiences is the same, the acceleration due to gravity, -9.81 m/s^2.
The coconut which is dropped from the top of the cliff "falls" 20 meters, so lets call that a change in distance of -20 m, and the coconut which is thrown upward travels upward 20 meters.

Solve these equations by first solving for the time, then use this value to find the initial velocity.

As I worked it, I found the time in the air before collision was about 2.02 second and the initial upward velocity imparted to the coconut was about 19.8 m/s.
But the final answer is less important than understanding how to solve it.

Answer 2

I agree with the first 2 answers. odu83's mistake: In his ascending coconut analysis, he used the time calculated as the time it takes for the falling coconut to reach 20 m, but used 40 m as the distance.

19.7 m/s is what I got - 19.8 is good too.

Answer 3

Mr Jeffy did notcompute the average velocity of the ascending coconut correctly


compute time of travel of falling coconut:

average v=d/t
d=vt
v=(iv+at)/2
iv=0
a=9.8m/s^2
d=20m
t=sqrt(40/9.8)s

now look at the ascending coconut

d=(ivt-at^2)/2
40=iv*sqrt(40/9.8)-9.8*(40/9.8)
80/sqrt(40/9.8)=iv
iv=39.6m/s

Answer 4

i counted its around 19.8 m/s..not sure if im right though