Solving Kinematics Problem: V_o & Max Height

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Homework Help Overview

The discussion revolves around a kinematics problem involving the calculation of initial velocity and maximum height related to an object in motion. Participants are attempting to analyze the situation based on given time intervals and equations of motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • One participant attempts to calculate the initial velocity and height using kinematic equations but expresses confusion over discrepancies between their results and expected answers. Others question the interpretation of time intervals and the wording of the problem.

Discussion Status

Participants are actively engaging with the problem, with some providing clarifications and humorous remarks. There is a recognition of potential misunderstandings regarding the timing of events in the problem, but no explicit consensus has been reached on the correct approach.

Contextual Notes

There appears to be a lack of clarity regarding the specific details of the problem setup, particularly concerning the timing of the events described. This has led to varying interpretations among participants.

DeltaForce
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Homework Statement
John climbs a tree to get a better view of the speaker at an outdoor graduation ceremony. Unfortunately, he leaves his binoculars behind. Marsha throws them up to John, but her strength is greater than her accuracy. The binoculars pass John's outstretched hand after 0.75s and again 1.28s later. How high is John?
Relevant Equations
v_o + at = v_f
v_o*t - 0.5gt^2 = y
For this problem I tried to find when the binoculars reaches the maximum height. So, (0.75 + 1.28)/2 = 1.015s. Using that information I can solve for the initial velocity. v_o = gt = 9.947 m/s.

Then using the initial velocity I can solve for the height of John using the 2nd kinematics equation.

9.947(0.75) - 0.5(9.8)(0.75^2) = 4.70m

I checked the answer and solution, but it was totally different from my answer. What did I do wrong?
 
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DeltaForce said:
How high is John?
Obviously we need more information to answer this question. What was he smoking before climbing the tree? :wink:
For this problem I tried to find when the binoculars reaches the maximum height. So, (0.75 + 1.28)/2 = 1.015s. Using that information I can solve for the initial velocity. v_o = gt = 9.947 m/s.

Then using the initial velocity I can solve for the height of John using the 2nd kinematics equation.

9.947(0.75) - 0.5(9.8)(0.75^2) = 4.70m

I checked the answer and solution, but it was totally different from my answer. What did I do wrong?
Check the wording of the question. The 1.28s is *after* the 0.75s first cross-over time... :smile:
 
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Hahahaha, I was sitting there wringing my brain out figuring out what I did wrong. Now I see. Thank you!
 
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Happy to help. I try to avoid climbing trees after, well, you know... o0)
 
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i dont get it
 
Hristijan said:
i dont get it
What exactly is it that you don't get?
 
kuruman said:
What exactly is it that you don't get?
Probably my witty humor. Not everybody gets me. :wink:
 
berkeman said:
Probably my witty humor. Not everybody gets me. :wink:
Ask whomever you think doesn't get your humor to climb a tree and get high, not necessarily in that order. Say anything and listen for the telltale giggles. On second thought, forget the tree.
 
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