Solving a Falling Pen Problem: Find the Roof Height

[james007]

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.

 

[rl.bhat]

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.

 

[JANm]

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

 

[rl.bhat]

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.

 

[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.

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

 

[rl.bhat]

Yes. It is correct. Now put the value of (t1 - t2) and find the value of t2.