I am having some trouble starting this problem:
A student is staring idly out her dormitory window when she sees a water balloon fall past. If the balloon takes 0.2 seconds to cross the 140 cm high window, from what height above the top of the window was it dropped?

Two steps - first, recognizing that the balloon is accelerating as it falls, it has a higher veocity as it passes the bottom of the window than it did at the top. So you will need to use:


{V_f}^2 - {V_i}^2= 2 ad


Here a = g = 9.8m/s^2,and d= 0.14m.

You also know the object takes 0.2 seconds to fall past the window. Using

\Delta V = \frac a t

lets you replace V_f in the first equation with V_i + \Delta V

Can you solve for V_i now?

5.1

5.1

Wrong. This problem is tricky, since for an object to take 0.2 seconds to fall 0.14 meters requiires that it start with an initial upward velocity (from a stranding start an object requires only 0.169 seconds to fall 140 cm). Hence the balloon did not start above the window at all. I suspect that OP mis-typed the question.

And why are you responding to a question that is almost three years old?

Are you sure ebaines? I'm getting something else... :(

Ok, this is what I used.

s = ut+ \frac12 at^2

So, I'm finding the initial speed of the water balloon at the top of the window panel.

1.4 = u(0.2) + \frac12 (9.8)(0.2^2)

1.4 - 0.196 = 0.2u

1.204 = 0.2 u

u = 6.02 ms^{-1}

Then, using v^2 = u^2 + 2as

(6.02^2) = 0 + 2(9.8)s

36.2404 = 19.6s

s = 1.849 m

And the time for a free falling body to make 140 cm is:

1.4 = 0t + \frac12 (9.8)t^2

t = \sqrt{\frac{1.4(2)}{9.8}} = 0.535 s

If I'm making a mistake, show me please.

Are you sure ebaines? I'm getting something else... :(


That'll teach me to be in a rush. Yes - you are correct. I made the mistake of converting 140 cm in my head to 0.14 meters, but of course it's actually 1.4 meters. Doh! I too get an answer of 1.849 meters.

That also reassures me, lol! The exams are in about two weeks for me, and I fear that I forget something :eek:

Thanks :)