Ball with maximum velocity on reaching the ground.

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Homework Help Overview

The problem involves three balls: one thrown upward to a height H, another thrown downward from that height, and a third dropped from the same height. The goal is to determine which ball has the maximum velocity upon hitting the ground.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the velocities of each ball upon impact, with some suggesting that the ball thrown downward will have a greater final velocity than the ball thrown upward. Others express confusion regarding the initial conditions of the balls and how they affect the final velocities.

Discussion Status

There appears to be a lack of consensus regarding the initial conditions of the balls, particularly whether one ball is thrown from height H or to height H. Some participants have attempted calculations based on different assumptions, while others have expressed confusion about the implications of these assumptions. The discussion has not reached a definitive conclusion, but various interpretations and calculations have been shared.

Contextual Notes

Participants are grappling with the clarity of the problem statement, particularly the phrasing regarding the direction of the throw for ball A. This ambiguity has led to differing interpretations and calculations among participants.

takando12
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Homework Statement


A ball A is thrown upward to a height H with a velocity of 20m/s, another ball B is thrown down from H with a velocity of 20m/s and a third ball C is just dropped from the same height. Find which ball has the maximum velocity when it hits the ground.

Homework Equations


v=u+gt , s=ut +1/2gt2

The Attempt at a Solution


A - The velocity on hitting the ground must be the same as the projected velocity. So I think it's 20 m/s
B - Since it starts with 20m/s from top ,and it's going to speed up because of gravity, it's reasonable to assume that it's final velocity will be greater than A's.
C-This is where I'm confused. Using the formula v=u+gt and u=0 . v=gt .So the final velocity of C is solely dependent on t. Taking the falling part of A's journey, u=0 and v= 20m/s( from first assumption) using v=u+gt. gt=20 . Is the time for the falling of A and C the same and so the answer is A=C<B ?How do i proceed?
 
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yesterday i told you the solution to this question. i think that you don't get that.
 
clear my point that, A is thrown from H or thrown from ground to H? if it is thrown from ground to H then, final velocity will be 20m/s, using formula we can get that H=20m(u^2/2g=h), by using (2gh)^.5 we can calculate final velocity of C, and v*v-u*u=2gH, you can calculate final velocity of B, no need to find time. please just put values and solve, and it think there should be no confusion
 
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Sagar Singh said:
clear my point that, A is thrown from H or thrown from ground to H? if it is thrown from ground to H then, final velocity will be 20m/s, using formula we can get that H=20m(u^2/2g=h), by using (2gh)^.5 we can calculate final velocity of C, and v*v-u*u=2gH, you can calculate final velocity of B, no need to find time. please just put values and solve, and it think there should be no confusion
There is no confusion at all. Firstly, the answer you said yesterday was that the velocity of both A and B on reaching the ground is 20m/s right? that is what is confusing.
using the method you yourself just gave.
A- H=20m v=20m/s
B- v2-u2=2gh
v2= 2*10*20 +400
v= 28.2 m/s
C- u=0m/s
v2-u2=2gh
v2= 2*10*20
v= 20m/s
it's A=C<B right? correct me if I am wrong.
 
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takando12 said:
There is no confusion at all. Firstly, the answer you said yesterday was that the velocity of both A and B on reaching the ground is 20m/s right? that is what is confusing.
using the method you yourself just gave.
A- H=20m v=20m/s
B- v2-u2=2gh
v2= 2*10*20 +400
v= 28.2 m/s
C- u=0m/s
v2-u2=2gh
v2= 2*10*20
v= 20m/s
it's A=C<B right? correct me if I am wrong.
and yes A is thrown from the ground to H.
 
takando12 said:
There is no confusion at all. Firstly, the answer you said yesterday was that the velocity of both A and B on reaching the ground is 20m/s right? that is what is confusing.
using the method you yourself just gave.
A- H=20m v=20m/s
B- v2-u2=2gh
v2= 2*10*20 +400
v= 28.2 m/s
C- u=0m/s
v2-u2=2gh
v2= 2*10*20
v= 20m/s
it's A=C<B right? correct me if I am wrong.
i mis under stood the question, i think you told me that A is thrown from height H, but today you said that A is thrown TO height H, there is much difference in between them. if TO is used than this is the correct answer, if From is used than yesterday post is correct answer
 
the question is solved
 
-
Sagar Singh said:
the question is solved
It's a mistake on my part. I never knew the question was not clear.I had this answer in the old question, it was on my mind from the beginning but then you said the opposite and i was completely confused the whole of yesterday. I re posted it today with the edit.
Well thank you Mr.Sagar and yes it has been solved. CHEERS!
 
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takando12 said:
-
It's a mistake on my part. I never knew the question was not clear.I had this answer in the old question, it was on my mind from the beginning but then you said the opposite and i was completely confused the whole of yesterday. I re posted it today with the edit.
Well thank you Mr.Sagar and yes it has been solved. CHEERS!
and btw it's not "brother" like you called me yesterday. :smile: I'm a girl.
 
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  • #10
takando12 said:
and btw it's not "brother" like you called me yesterday. :smile: I'm a girl.
ohh sorry, *sister*
 

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