Finding Maximum Height of Two Simultaneously Launched Balls

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Homework Help Overview

The problem involves two rubber balls: one is shot straight up from the ground with an initial speed \( v_0 \), while the other is dropped from a height \( h \) directly above the first ball. The goal is to find the maximum height \( h_{max} \) at which a collision occurs before the first ball returns to the ground.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the definition and significance of \( h_{max} \) and its relation to the collision of the two balls. There are attempts to derive equations related to the motion of the balls, with some participants expressing confusion about missing terms in their calculations.

Discussion Status

The discussion includes various interpretations of the problem, with some participants questioning the clarity of the original poster's question. Guidance has been offered regarding the relationship between the time of flight and the conditions for collision, but no consensus has been reached on the specific calculations.

Contextual Notes

Some participants mention the challenges posed by the resources they are using, indicating that there may be constraints in understanding the problem fully. The original poster has expressed difficulty in identifying missing components in their solution attempt.

EHogeberg
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This problem has been giving me problems for the past two days and I can't figure it out :( please help. this is my first ever physics class.

Homework Statement


A rubber ball is shot straight up from the ground with speed v_0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest.

Known:
V_0 and Height of original ball drop
y_coll= h- ((gh^2)/(2v_0)^2)
find:
h_max

Homework Equations



Vx= Vxi+ax (tf-ti)
Xf=Xi + Vxi + 1/2 ax (tf-ti)^2
Vxf^2= Vxi^2+ 2ax (Xf-Xi)

The Attempt at a Solution



I got to the point where I found what I thought was h_max and got

hmax= (V_0^2/g)- 1/2*V_0^2

the website I am working on told me I am missing a term but I can't figure out what I am missing. :(
 
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The question is, what is the question? You didn't state what it is you're trying to find!
 
how to find h_max
 
EHogeberg said:
how to find h_max

Yes, but what is h_max supposed to represent?
 
the maximum value of h for which a collision occurs before the first ball falls back to the ground?
 
Last edited:
EHogeberg said:
the maximum value of h for which a collision occurs before the first ball falls back to the ground?
I would have never guessed that.
 
If a ball is shot vertically upward, with initial velocity, v0, how long does it take the ball to return to its initial location?
 
I know this book and website combination is not the most helpful thing in the world. Whaat I have done so far is pluged in Hmax for the Xf of the second kinematic equation and when I solved it I ended up getting v_0^2/g-1/2 v_o^2 which got me to where it said I was close but I was either missing a term or it was incorrect and I can't figure out what's wrong so I guess my question should be...what am I doing wrong
 
SammyS said:
If a ball is shot vertically upward, with initial velocity, v0, how long does it take the ball to return to its initial location?

sorry I didnt see that question...I am not sure the only thing I can think of is time final-time initial but that wouldn't help me. I really don't knowfrom what I can think of it would equal the velocity over the gravity
 
  • #10
EHogeberg said:
sorry I didnt see that question...I am not sure the only thing I can think of is time final-time initial but that wouldn't help me. I really don't know

from what I can think of it would equal the velocity over the gravity
If it wouldn't help, I wouldn't have asked it. It will definitely help to work that out.
 
  • #11
what I meant to say is my answer wasnt going to help me sorry for the unclarity
 
  • #12
I finnaly figured it out...thank you guys for the guidence
 
  • #13
Good. The collision should occur just before the first rubber gets back to ground level.
 
  • #14
yeah that's what I got
 

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