Free Falling Objects: Find Initial Velocity of Q

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Discussion Overview

The discussion revolves around a physics problem involving two stones, one released from a height and the other launched downward, with the goal of finding the initial velocity of the second stone given that both stones reach the ground simultaneously. The scope includes mathematical reasoning and problem-solving related to kinematics.

Discussion Character

  • Mathematical reasoning, Homework-related, Debate/contested

Main Points Raised

  • One participant states the initial conditions and attempts to calculate the initial velocity of stone Q using kinematic equations, arriving at a time of 5 seconds for the first stone's fall.
  • Another participant confirms the use of correct equations and emphasizes the need to determine the correct time of flight for the second stone, prompting further clarification.
  • A participant calculates the distance stone Q falls as 102.9m, based on the height of the cliff minus the distance fallen by stone P.
  • Another participant challenges this interpretation, asserting that stone Q is released from the same height as stone P and must therefore fall the full height but in a shorter time, leading to a need for recalculating the time of flight for stone Q.
  • A participant expresses confusion about the variables involved, particularly regarding the time of flight.
  • One participant reiterates that the time of flight for the second stone must be shorter since it is released after the first stone has fallen a distance of 19.6m.
  • A final participant acknowledges understanding after the clarification about the time of flight.

Areas of Agreement / Disagreement

Participants do not reach consensus on the interpretation of the problem initially, with some misunderstanding the time of flight for the second stone. However, there is agreement on the necessity to find the correct time for stone Q to solve the problem.

Contextual Notes

There are unresolved assumptions regarding the interpretation of the problem and the calculations of time of flight for both stones, which may affect the final results.

stupif
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1. stone p is releasedfrom a cliff of height 122.5m above the ground. after falling a distance of 19.6m, another stone Q is lauched straight downward.if given that p and Q reached the ground same time, find the initial velocity of Q( answer :26.1m/s

i using s=ut +1/2a[t][2] to find out the t, t= 5s
time of p=time of Q
i using s=ut +1/2a[t][2] to find out the initial velocity of u by change the S.
my answer is -3.92m/s
 
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Hi stupif. You're using the correct equations and your figure of 5 seconds is correct for the time of flight of the first stone.

The key to getting the correct answer is to figure out the correct time of flight for the second stone. So tell us what number you're using for this quantity and how you arrived at it.
 
then i used 122.5m - 19.6m=102.9m, so i using 102.9m as stoneQ's distance
 
stupif said:
then i used 122.5m - 19.6m=102.9m, so i using 102.9m as stoneQ's distance
No you're interpreting the question incorrectly. The second stone is released from the same height as the first one (that is, 122.5m). This means that the second stone falls the same distance but does so in a shorter time period. If you work out what this shorter time period is then you can easily work out the required initial velocity. (the books answer is correct btw).
 
then s= same, t=same, a=same, just u is variable. how to calculate??
 
No, "t" is NOT the same. The second stone is released AFTER the first stone has already fallen 19.6m, yet they arrive at the ground at the same time! So the time of flight of the second stone MUST be smaller. So that's the key to the problem, find the time of flight for the second stone and then you can easily find "u".

How long does it take for the first stone to fall 19.6m?

Use the above to find the time of flight for the second stone.
 
thank you...i got it...
 

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