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

Click For Summary
SUMMARY

The discussion centers on solving a projectile motion problem involving two heights, h and H, reached at different times, t1 and t2, with the relationship t2 = 3t1. The equations of motion used are h = ut1 - (1/2)gt1² and H = ut2 - (1/2)gt2². Participants derive the relationship h = (5/9)H, which conflicts with the book's answer of (3/4)H. The conversation emphasizes the importance of correctly applying SUVAT equations and understanding the relationships between variables in projectile motion.

PREREQUISITES
  • Understanding of projectile motion and kinematics
  • Familiarity with SUVAT equations
  • Knowledge of gravitational acceleration (g)
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of SUVAT equations in detail
  • Explore the concept of free fall and its equations
  • Investigate the relationship between initial velocity and maximum height in projectile motion
  • Practice solving problems involving multiple heights and times in projectile motion
USEFUL FOR

Students and educators in physics, particularly those focusing on kinematics and projectile motion, as well as anyone preparing for exams involving these concepts.

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   Reactions: 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?
 

Similar threads

Replies
40
Views
2K
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
26K
Replies
1
Views
2K