How Far Below the Cliff Top Do the Two Balls Cross Paths?

[afcwestwarrior]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

What equation should I use.

 

[Astronuc]

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.

 

[m2003]

To find the distance at which the two balls cross paths, you can use the equation for displacement:

d = (V0t + 1/2at^2)

Where d is the distance, V0 is the initial velocity, a is the acceleration (in this case, due to gravity), and t is the time.

Since the initial velocity for both balls is the same, you can set V0 to be equal for both equations:

V0 = 0 for the first ball (since it is dropped from rest)
V0 = initial velocity for the second ball (which is the same as the final velocity for the first ball)

You can also set the acceleration (a) to be the same for both equations, since the only force acting on both balls is gravity.

Now, you can set up two equations:

d1 = 1/2at^2 (for the first ball)
d2 = V0t + 1/2at^2 (for the second ball)

Since you want to find the distance at which the two balls cross paths, you can set d1 = d2 and solve for t.

1/2at^2 = V0t + 1/2at^2

V0t = 0

t = 0

This means that the two balls will cross paths at the same time (t=0). To find the distance at which this happens, you can plug in t=0 into either of the equations (since they are both equal now):

d1 = 1/2at^2 = 1/2a(0)^2 = 0

d2 = V0t + 1/2at^2 = V0(0) + 1/2a(0)^2 = 0

Therefore, the two balls will cross paths at a distance of 0m from the top of the cliff. This makes sense, since they are both dropped and thrown from the same height and at the same initial velocity, so they will cross paths at the same height.