# I must find distance.

1. Sep 2, 2007

### afcwestwarrior

1. The problem statement, all variables and given/known data A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upward at the same instant that the 1st ball is dropped. The initial speed of the second ball is exactly the same as that which the 1st ball eventually hits the ground. In the absence of air resistance, the motions of the balls are just reverse of each other. Determine how far below the top of the cliff the balls cross paths.
Height= 24 m

2. Relevant equations
y=1/2 (Vo + V) t
y= Vot + 1/2at
V^2= Vo^2 + 2ay or y=V^2 - Vo^2/2a

3. The attempt at a solution

What equation should I use.

2. Sep 2, 2007

### Staff: Mentor

Well first, determine the velocity of the 1st ball.

It drops 24 m with an acceleration of -9.81 m/s2.

The initial velocity of a ball that is 'dropped' is zero.

The 2nd ball travels vertically with an initial velocity and decelerates.

See -
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#ffall

The first ball is increasing speed and the second ball starts with a velocity and decreases in speed. One travels a distance h and the other a distance 24-h in the same amount of time.

Last edited: Sep 2, 2007
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