Solving the Falling Ball Problem: t2=2t1?

[princesspriya]

Homework Statement

two students are standing on a fire escape, one twice as high as the other. simultaneously, each drops a ball. if the first ball strikes the ground at time t1 when willt he second ball strike the ground?
A)t2=4t1
B)t2=5/4t1
C)t2=2t1
D)t2=squareroot of 2 times t1

Homework Equations

The Attempt at a Solution

i tried plugging in numbers and well its two times the height so wouldn't it be C??

 

[blochwave]

Yay I get to use my skydiving experience! When jumping out of an airplane I know it'll take roughly 10 seconds to fall the first 1000 feet, then roughly 5 seconds for every subsequent 1000 feet. So 10 seconds for 1000 feet, 15 seconds for 2000 feet(pretty gross estimation, but it overestimates how far you've fallen usually which is better than underestimating for obvious reasons!)

So think about the equation you should be using here

d=-1/2*g*t^2

What you've said in thinking it's C is that if I have a ball that falls a distance D1 in T1 seconds then if T2=2*T1, then D2=2*D1

is that true?

 

[princesspriya]

wen i tried pluging in numbers i kept the initial velocity, aceleration the same and made one distance x and the other one 2x and then i tried finding the final velocity. then using one of the formulas i found how much time it would take and compared them. so the Vi and A is the same but the D and Vf are different.

 

[blochwave]

You'll confuse yourself trying to plug in numbers

so D=-1/2*g*T^2, 1/2 and g are constants

Now we're saying it falls twice D in time T2, so 2*D=-1/2*g*T2^2, and you have said that T2=2*T, or that's what choice C says anyways. So what you're saying is 2*D=-1/2*g*(2*T)^2

You KNOW that D=-1/2*g*T^2 from our very first step, so if you solve the bolded equation for D, do you get the same thing?

 

[princesspriya]

yea the both equation do equal out. but how would that prove that the time would be 2 times if the distance is 2x?