Two balls are thrown vertically upward, one with an initial speed twic

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Homework Help Overview

The problem involves two balls thrown vertically upward, one with an initial speed that is twice that of the other. Participants are discussing the heights reached by each ball and the implications of their initial speeds on these heights.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are exploring different kinematic equations to determine the relationship between initial speed and maximum height. There are conflicting interpretations regarding whether the ball with the greater initial speed reaches twice or four times the height of the other. Some participants question the validity of the results obtained from the equations used.

Discussion Status

The discussion is ongoing, with participants providing insights into the equations and their applications. Some have pointed out potential errors in reasoning, while others are encouraging further exploration of the concepts involved. There is a recognition of the complexity of projectile motion, and some participants suggest calculating the time to reach the highest point as a helpful approach.

Contextual Notes

Participants have noted typographical errors in the equations presented and are discussing the implications of these errors on the results. There is also an acknowledgment of the common challenges faced in understanding projectile motion problems.

cmkc109
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Homework Statement



Two balls are thrown vertically upward, one with an initial speed twice that of the other. The ball with the greater initial speed will reach a height

- twice that of the other
- 4 times that of the other

Homework Equations


dy = vt + 1/2 at^2
[if i use this equation, i will get twice of the other]

vf^2 = v1^2 + 2ad
-v1^2 = 2ad
[if i use the equation, i will get 4 times of the other]

The Attempt at a Solution



I get both 2 times and 4 times..
 
Last edited:
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cmkc109 said:

Homework Equations


dy = v + 1/2 at^2
[if i use this equation, i will get twice of the other]

This is incorrect. You might want to show here how you get this result.
 
voko said:
This is incorrect. You might want to show here how you get this result.

i made a typo
it is dy = vi t + 1/2 at^2

this is one of the kinematic equation
 
It seems to me that projectile motion is difficult for many because every second post is about it but that's ok.
 
cmkc109 said:
i made a typo
it is dy = vi t + 1/2 at^2

this is one of the kinematic equation

It is indeed a valid equation, but your result is still incorrect.
 
lep11 said:
It seems to me that projectile motion is difficult for many because every second post is about it but that's ok.

Galileo created the whole science called "physics" by considering projectile motion, so it should not be surprising that it is a hot topic in teaching of this science. That's why it is so frequent here, not because it is difficult for many. Well, some problems may be surprisingly difficult.
 
voko said:
Galileo created the whole science called "physics" by considering projectile motion, so it should not be surprising that it is a hot topic in teaching of this science. That's why it is so frequent here, not because it is difficult for many. Well, some problems may be surprisingly difficult.
Yes, but in general projectile motion problems are easy because if you understand the concept you can solve the problem using only a couple of kinematic equations.
 
voko said:
It is indeed a valid equation, but your result is still incorrect.


why is it incorrect? can u explain why?
 
cmkc109 said:
why is it incorrect? can u explain why?

The correct result is 4 times. I cannot explain why you are not getting the correct result using this method, you have not shown your working.
 
  • #10
voko said:
The correct result is 4 times. I cannot explain why you are not getting the correct result using this method, you have not shown your working.


i did not do any work. it's simply because dy - at^2 = vi t
so there is no square sign
 
  • #11
What about t? Is it the same in both cases?
 
  • #12
voko said:
What about t? Is it the same in both cases?

no.. the ball with higher velocity has shorter time..o...i get it loll
using vf^2 = vi^2 + 2ad , a is constant for both cases
but if u use the other equation, t varies so the ans is not correct !
thhxx
 
  • #13
It might be helpful for you to compute the time it takes to reach highest point in both cases.
 

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