Solve 1D Kinematics Homework: Collision of 2 Rubber Balls

[Copyright]

Homework Statement

A rubber ball is shot straight up from the ground with speed v_o. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest.

At what height above the ground do the balls collide? Your answer will be a symbolic expressions in terms of v_o and g.

Homework Equations

v=v0+at
x=x0+v0t+(1/2)at2
v2=v20+2aΔx

The Attempt at a Solution

For the first ball, I have $$v_f = \sqrt{v^2_o+2g}$$
For the second ball (which I think I have wrong), I have $$v_f = \sqrt{2(g)(-h)}$$

Unfortunately I'm not really sure where to go from here, or if I'm even headed in the right direction in the first place. Can someone please help?


Thanks.

 

[ehild]

For both balls, write up the time-dependence of height in terms of the initial position and velocity (your second equation) . When they meet, their heights are equal: that means an equation for the time. Solve, and determine the height with the result.

ehild

 

[Copyright]

First ball y=vot + .5(g)(t^2) second ball y=h+.5(g)(t^2). Am I solving for the wrong variable (you mentioned time)? Also does the derivative come into play? Sorry for all the questions, I've been up for a long time .EDIT: Do you set them equal to each other?
h+.5gt^2 = vot + .5gt^2
h = vot
t = h/vo

??

 

[ehild]

Are you sure that the sign of g is correct? If it is taken positive, will the balls fall to the ground at all?

ehild

 

[rwisz]

First, let's define some variables for easier understanding:

Vo = initial velocity of the first ball (upward)
H = initial height of the second ball above the ground
Vf = final velocity of both balls before the collision
T = time it takes for the balls to collide

Using the kinematic equation v=v0+at, we can find the final velocity of the first ball before the collision:

Vf = Vo + gt

To find the final velocity of the second ball, we can use the equation v2=v20+2aΔx, where Δx is the displacement of the second ball before the collision.

Δx = H (since the ball is dropped from rest)

Substituting this into the equation, we get:

Vf = √(0 + 2gH)

Now, we know that the final velocities of both balls must be equal for them to collide. So we can equate the two expressions for Vf:

Vo + gt = √(2gH)

Solving for t, we get:

t = (√(2gH) - Vo) / g

Now, using the equation x=x0+v0t+(1/2)at2, we can find the height at which the balls collide by substituting the time t we just found and the initial height H of the second ball:

X = 0 + 0 + (1/2)g[(√(2gH) - Vo) / g]2 + H

Simplifying, we get:

X = H + (√(2gH) - Vo)^2 / 2g

Therefore, the height at which the balls collide is given by the expression:

H + (√(2gH) - Vo)^2 / 2g

This is a symbolic expression in terms of Vo and g, as requested in the homework statement.