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

AI Thread Summary
The discussion revolves around a projectile motion problem involving two heights, h and H, and their respective times, t1 and t2, with a given ratio of t1 to t2 as 1/3. Participants explore the relationships between initial speed u, height H, and the equations governing motion under gravity, particularly focusing on the equations h=ut1-1/2gt1^2 and H=ut2-1/2gt2^2. There is confusion regarding the derived relationship h=(5/9)H, which does not match the book's answer of (3/4)H, prompting further investigation into the correct application of SUVAT equations. Suggestions are made to consider the problem from the perspective of an object in free fall to clarify the calculations. The conversation emphasizes the importance of correctly manipulating simultaneous equations to find the desired relationships.
rudransh verma
Gold Member
Messages
1,067
Reaction score
96
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.
 
Physics news on Phys.org
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##
 
  • Like
Likes BvU
@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?
 
Back
Top