Solving a Falling Pen Problem: Find the Roof Height

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Homework Help Overview

The problem involves a pen falling from a roof, with the specific detail that it takes 1.5 seconds to fall the last quarter of the total height. The objective is to determine the height of the roof from the ground, utilizing kinematic equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to clarify the initial conditions, particularly the initial velocity of the pen. Some participants suggest using kinematic equations to relate the total height and the time taken to fall specific distances. Others explore the relationships between the times taken to fall different segments of the height.

Discussion Status

Participants are actively engaging with the problem, offering various equations and methods to relate the different time intervals involved in the pen's fall. There is a focus on deriving relationships between the variables, although no consensus on a final approach has been reached yet.

Contextual Notes

There is some ambiguity regarding the initial velocity of the pen, as the problem does not specify whether it falls from rest. Additionally, the time taken to fall the last quarter of the height is a key piece of information that participants are working to incorporate into their equations.

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Homework Statement


A pen falls from the top of a roof. We know that it takes 1.5 seconds to fall the last 1/4 distance. Find the height of the roof from the ground.


Homework Equations


x=x_0 + v_0 * t + 1/2 gt^2
v=v_0 + gt



The Attempt at a Solution


Since the problem does not say that the pen falls from rest, I don't know what v_0 should be. Also, how can I make use of the time that it travels the last 1/4 distance.
Any help please.
 
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A pen falls from the top of a roof. So its initial velocity = 0. Let t1 be the time to reach the ground ( h ) and t2 be the time to cover the distance 3h/4. Write down two equations;
h = 1/2*g*t1^2...(1)
3h/4 = 1/2*g*t2^2 ...(2)
You can rewrite the two equation in terms of t1 and t2.
Since t1 - t2 is given, you can find h.
 
rl.bhat said:
h = 1/2*g*t1^2...(1)
3h/4 = 1/2*g*t2^2 ...(2)
Since t1 - t2 is given, you can find h.
hello rl.bhat
(1)-(2) gives h/4=1/2*g(t1-t2)*(t1+t2) and so t1+t2 can be easily calculated but what from there?
greetings Janm
 
Divide (1) by (2) you get
4/3 = t1^2/t2^2
2/(3)^1/2 = t1/t2
(2/1.732) - 1 = (t1/t2) - 1
Solve this equation to find t2.
 
rl.bhat said:
Divide (1) by (2) you get
4/3 = t1^2/t2^2
2/(3)^1/2 = t1/t2
(2/1.732) - 1 = (t1/t2) - 1
Solve this equation to find t2.
Hello rl.bhat
For sqrt(3) I tend to use 26/15. So your third equation becomes:
30/26=15/13=t1/t2
I am getting a hunch why you write the fourth equation, which becomes:
2/13=(t1-t2)/t2
is that correct?
greetings Janm
 
Yes. It is correct. Now put the value of (t1 - t2) and find the value of t2.
 

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