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

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SUMMARY

In the discussion, two balls are thrown vertically upward, one with an initial speed twice that of the other. The correct conclusion is that the ball with the greater initial speed will reach a height four times that of the other ball, as derived from the kinematic equation vf^2 = vi^2 + 2ad. The participants clarify that while both dy = vi t + 1/2 at^2 and vf^2 = vi^2 + 2ad are valid equations, the former leads to an incorrect conclusion due to the variable time factor. Understanding projectile motion and the application of kinematic equations is essential for solving such problems.

PREREQUISITES
  • Understanding of kinematic equations, specifically dy = vi t + 1/2 at^2 and vf^2 = vi^2 + 2ad
  • Basic knowledge of projectile motion principles
  • Familiarity with the concept of initial velocity and its impact on motion
  • Ability to analyze and differentiate between time variables in motion equations
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn how to compute the time taken to reach the highest point in projectile motion
  • Explore the effects of varying initial velocities on the trajectory of projectiles
  • Practice solving projectile motion problems to reinforce understanding of concepts
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in mastering projectile motion concepts and problem-solving techniques.

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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