Calculate the change in time between two throws

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Homework Help Overview

The discussion revolves around the timing of two balls thrown by a boy, specifically questioning the notion that both balls have the same final time 't' despite being thrown at different instances. The subject area involves kinematics and time intervals in projectile motion.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants express confusion regarding the equality of the final time 't' for both balls, questioning the assumptions behind this concept. There are attempts to clarify the relationship between the times of the two throws and the implications of their respective timing mechanisms.

Discussion Status

The discussion is ongoing, with participants exploring the reasoning behind the timing of the throws. Some guidance has been offered regarding the relationship between the two time measurements, but no consensus has been reached on the underlying assumptions.

Contextual Notes

Participants are grappling with the implications of the timing setup, particularly how the time intervals are defined and measured for each ball. There is an emphasis on understanding the difference between the two time variables involved.

Altairs
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Homework Statement


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Homework Equations



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The Attempt at a Solution



The confusion is that why will the final time of the two balls be the same i.e. 't'...The time taken by the two balls can never be the same...Am confused at the same final time 't'..Please explain.
 

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Hi Altairs! :smile:
Altairs said:
The confusion is that why will the final time of the two balls be the same i.e. 't'...The time taken by the two balls can never be the same...Am confused at the same final time 't'..Please explain.

Because the boy throws the first ball at t = 0, and the second ball at t = ∆t.

That's why (t - ∆t) keeps appearing in the equations! :wink:
 
My pooint is why the same 't' ?
 
Altairs said:
My pooint is why the same 't' ?

It's as if the boy has two stop-watches, one for each ball, each starting at zero when that ball is thrown.

The first watch measures time t, starting at t = 0, and so it shows t = ∆t when the second ball is thrown.

The second watch measures time u, starting at u = 0, which is the same as t = ∆t.

From then on, the second watch is always ∆t slower than the first watch … in other words, u = t - ∆t.

The two formulas do not use the same t … they use t and u …

but you can convert u into t. :smile:
 

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