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

• rudransh verma
In summary, the conversation discusses an object with initial speed u that first attains height h and then H at times t1 and t2 respectively. By substituting t2 with 3t1, the relationship between H and u can be found to be u^2 = 2gH. The conversation also mentions using another SUVAT equation to find the relationship between u and the vertical velocity at H. The resulting equation is h = (5/9)H, which is different from the book's answer of (3/4)H.
rudransh verma
Gold Member
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.

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## ?

[ ah, haru was faster !

##\ ##

BvU said:
What is the relationship between H and u ?
Maybe we can use ##u^2=2gH##

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?

haruspex said:
What does the book get?
(3/4)H

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?

## 1. What is a projectile problem?

A projectile problem is a type of physics problem that involves calculating the motion of an object that is launched or thrown into the air, also known as a projectile. These problems typically involve finding the object's height, distance, and time of flight.

## 2. How is time to reach 1/3 of the maximum height calculated?

The time to reach 1/3 of the maximum height can be calculated by dividing the total time of flight by 3. This is because the time to reach the maximum height is half of the total time, and 1/3 of half is equal to 1/6 of the total time. Therefore, to find the time to reach 1/3 of the maximum height, you can simply divide the total time by 3.

## 3. Why is it important to calculate the time to reach 1/3 of the maximum height?

Calculating the time to reach 1/3 of the maximum height is important because it can help determine the overall trajectory of the projectile. This information is useful in predicting where the object will land and can be used to adjust the launch angle or velocity to achieve a desired outcome.

## 4. What factors affect the time to reach 1/3 of the maximum height?

The time to reach 1/3 of the maximum height is affected by the initial velocity, launch angle, and acceleration due to gravity. The higher the initial velocity and launch angle, the shorter the time to reach 1/3 of the maximum height. Conversely, a higher acceleration due to gravity will result in a longer time to reach 1/3 of the maximum height.

## 5. Can the time to reach 1/3 of the maximum height be greater than the total time of flight?

No, the time to reach 1/3 of the maximum height cannot be greater than the total time of flight. This is because the object must reach its maximum height before falling back down to the ground, and the time to reach 1/3 of the maximum height is always a fraction of the total time of flight.

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