Spiderman falls off a building... [physics]

LuisRodg · 2008-01-13 12:44:08

Spiderman steps from the top of a tall building. He falls freely from rest to the ground a distance of h. He falls a distance of h/4 in the last interval of time of 1.2 s of his fall.

What is h?

I have spent the last hour trying to solve this and for the love of god i cant solve it? Anyone guide me in how to solve it?

John E. Franklin · 2008-01-13 18:46:17

I'm still working on it, but I'll give you my preliminary answer in case you are waiting:
The left side of equation for distance is 3/4 of fall.
I multiply this by reciprocal or 4/3 to reach 100% of fall on right side of equation 1.2 seconds later.
4.9 t^2 (4/3) = 4.9 (t + 1.2)^2
I don't know if this is right yet...

John E. Franklin · 2008-01-13 19:08:59

I used quadratic equation to solve for t at 3/4 distance down point and got:
t = 6/5 ( 3 + 2√3)
Answer = 393.109608~ meters, if I'm right??

I decided to check my answer in different units of time.
I made up a new unit of time called a Jecond instead of a Second.
And the Jecond is 12/5 times longer than a second.
Then g = 9.8 m/s^2 changes to 56.448 m/Jecond^2
Redoing the quadratic equation resulted in easier numbers:
(56.448/2)(4/3)t^2 = (56.448/2)(t+1/2)^2
The t + 1/2 is time at hitting the ground.
1/2 of a Jecond is 1.2 seconds.
It works out to the same answer.
Thought you might like that !!

Last edited by John E. Franklin (2008-01-13 20:19:29)

mathsyperson · 2008-01-14 01:11:51

John's method is brilliant.

I didn't understand it at first, so tried to do it myself in a different way.
I split it into two parts, considering first the initial h metres of his fall.

We know the initial velocity, distance and acceleration of that fall, so we can work out the final velocity of it, in terms of h.

Then using that value as the initial velocity of the next part, along with the time and acceleration that are given, we can work out the distance travelled in terms of h.

But we already know that that distance is h/4, so then we equate the two distances and see which h works.

It got horribly complicated very quickly and I heartily recommend John's method over mine.

One thing though: Spiderman travels h metres, and then another h/4.
That means that John's equation should read:
4.9 t^2 (5/4) = 4.9 (t + 1.2)^2

Edit: Ignore me.

Daniel123 · 2008-01-14 04:09:23

One thing though: Spiderman travels h metres, and then another h/4.
That means that John's equation should read:
4.9 t^2 (5/4) = 4.9 (t + 1.2)^2

Does he? I read the question to mean that he falls a total distance of h, and the last quarter of this distance is travelled in 1.2s?

Daniel123 · 2008-01-14 04:28:51

My method:

Work out his velocity after falling 3/4h which, using v² = u² + 2as, is √(14.7h)

Now create an equation which you can solve for h, using s = ut + ½ at², with h/4 for s, 1.2 for t and √(14.7h) for u. Then just some ugly algebra to do.

mathsyperson · 2008-01-14 05:53:46

Oh, you're right. I should have read the question more carefully.
Your method is the same one that I was describing earlier. It's certainly possible to get the answer that way, but I like John's method a lot more.

Daniel123 · 2008-01-14 05:58:57

And I should have read your method more carefully!
Yes I agree, it's very clever.

Math Is Fun Forum

#1 2008-01-13 12:44:08

Spiderman falls off a building... [physics]

#2 2008-01-13 18:46:17

Re: Spiderman falls off a building... [physics]

#3 2008-01-13 19:08:59

Re: Spiderman falls off a building... [physics]

#4 2008-01-14 01:11:51

Re: Spiderman falls off a building... [physics]

#5 2008-01-14 04:09:23

Re: Spiderman falls off a building... [physics]

#6 2008-01-14 04:28:51

Re: Spiderman falls off a building... [physics]

#7 2008-01-14 05:53:46

Re: Spiderman falls off a building... [physics]

#8 2008-01-14 05:58:57

Re: Spiderman falls off a building... [physics]

Board footer