A Projectile problem -- Time to reach 1/3 of the max height

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Homework Help Overview

The discussion revolves around a projectile motion problem, specifically focusing on the time it takes to reach one-third of the maximum height. Participants are analyzing the relationships between initial speed, height, and time using kinematic equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to relate the heights attained at different times using kinematic equations. Questions about the relationships between variables such as height (H), initial speed (u), and time (t1, t2) are raised. Some participants suggest drawing graphs to clarify the problem.

Discussion Status

The discussion is active, with participants exploring various equations and relationships. Some guidance has been offered regarding the manipulation of equations and the importance of maintaining certain variables. Multiple interpretations of the problem are being considered, particularly regarding the expected answer compared to a reference book.

Contextual Notes

There is mention of a discrepancy between participants' findings and the solution provided in a reference book, which suggests a different relationship between the heights. The problem is framed within the constraints of homework rules, emphasizing the need for exploration rather than direct solutions.

rudransh verma
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Homework Statement
A particle is projected vertically upward and it attains max height H. If the ratio of times to attain height h(h<H) is 1/3, find h.
Relevant Equations
##s=ut+\frac12at^2##
Assuming it’s one body whose initial speed is u. First it attains height h then H. t1 and t2 are two times at which they attain h and H.
##h=ut1-\frac12gt1^2##
##H=ut2-\frac12gt2^2##
##\frac {t1}{t2}=1/3## Replacing t2 with 3t1, I am stuck.
 
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You have not used a certain fact. Can you spot it?
 
Make a drawing of height as a function of time. Is the question clear now ?
I mean iss the mathematical problem clear now ?

What is the relationship between ##H## and ##u## ?

[[edit] ah, haru was faster !

##\ ##
 
BvU said:
What is the relationship between H and u ?
Maybe we can use ##u^2=2gH##
 
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@BvU I got based on ##h=ut1-\frac12gt1^2## , ##H=ut2-\frac12gt2^2## , ##u^2=2gH## and ##t2=3t1##
##\frac{u^2}{2g}=3ut1-\frac92gt1^2## and
##h=ut1-\frac12gt1^2##
But then ?
 
Perhaps it's time to invoke another relevant SUVAT equation ... how about ##u## and the vertical velocity at ##H## -- what is their relationship ?

##\ ##
 
rudransh verma said:
@BvU I got based on ##h=ut1-\frac12gt1^2## , ##H=ut2-\frac12gt2^2## , ##u^2=2gH## and ##t2=3t1##
##\frac{u^2}{2g}=3ut1-\frac92gt1^2## and
##h=ut1-\frac12gt1^2##
But then ?
When manipulating simultaneous equations, you need to keep in mind which variables should appear in the final equation. The given variable is H, so don't try to eliminate it. The ones to be eliminated are t1, t2 and u.
 
BvU said:
Perhaps it's time to invoke another relevant SUVAT equation ... how about ##u## and the vertical velocity at ##H## -- what is their relationship ?

##\ ##
I am getting ##h=(5/9)H## but it’s not the same in book.
 
rudransh verma said:
I am getting ##h=(5/9)H## but it’s not the same in book.
I get the same.
The easiest way is to reverse it and consider an object in free fall from rest. If it falls distance y in time t, how far does it fall in time 2t? What about time 2t/3?

What does the book get?
 
  • #10
haruspex said:
What does the book get?
(3/4)H
 
  • #11
rudransh verma said:
(3/4)H
That would be the answer if the time were a half, not a third.
Did you try the way I mentioned in post #9?
 

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