While sitting on a tree branch 10.0 m above the ground, you drop a chesnut. When the chestnut has fallen 2.5 m, you throw a second one straight down. What initial speed must you give the 2nd chestnut if they are both to reach the ground at the same time?
Other variables given: a = -9.81 m/s^2
Constant acceleration equations of motion:
(all of the (f)'s and (i)'s indicate just final and initial - not multipliers)
Formula 1: v(f)^2 = v(i)^2 + 2(a)[x(f) - x(i)]
Formula 2: x(f) = x(i) + v(f)(t) + (1/2)(a)(t)^2
and maybe: x(f) = x(i) + (1/2)[v(i) + v(f)]t
The Attempt at a Solution
I thought I would find the velocity of the C1 (first chestnut) after falling 2.5 m, and then designate it v(i). Then I would make the two times equal to eachother and in turn manipulate to find the required v(i) of C2.
Here is how I found the velocity of C1 after falling 2.5 m:
Using formula 1:
v(f)^2 = 2(-9.81)[7.5-10]
v(f) = -7.0 m/s
Now I have the following variables for the second part:
v(i) = -7.0 m/s
x(i) = 7.5 m
x(f) = 0 m
v(i) = ??
x(i) = 10 m
x(f) = 0 m
So, I decided to use Formula 2, manipulate to solve for time for each, and then set each equation equal to eachother. However, it looks quite messy with the quadratic equations and such, so I'm guessing my approach must be wrong.
Maybe I should use the other equation I mentioned above instead? But then it would require a final velocity... I'm stumped!
Please help me out... it would be much appreciated!!