Solve 1D Kinematics Homework: Collision of 2 Rubber Balls

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The problem involves two rubber balls: one shot upwards with initial speed vo and another dropped from height h. To find the collision height, the equations of motion for both balls must be set equal, as their heights will be the same at the point of collision. The first ball's height is given by y = vot - (1/2)gt^2, while the second ball's height is y = h - (1/2)gt^2. By equating these two expressions and solving for the height, the result can be expressed symbolically in terms of vo and g. The discussion emphasizes the importance of correctly applying the equations of motion and the signs of acceleration due to gravity.
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Homework Statement


A rubber ball is shot straight up from the ground with speed vo. 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 vo 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.
 
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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
 
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 O_o.EDIT: Do you set them equal to each other?
h+.5gt^2 = vot + .5gt^2
h = vot
t = h/vo

??
 
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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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