# 2 objects falling at diff. times; must find init. vel. of 2nd if land at same time

by unigal13
Tags: diff, falling, init, land, objects, time, times
 P: 4 1. The problem statement, all variables and given/known data 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 2. Relevant equations 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 3. 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: C1: v(i) = -7.0 m/s x(i) = 7.5 m x(f) = 0 m C2: 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!!
 HW Helper P: 1,986 Welcome to PF. You know the time C1 takes to fall the 7.5 m height ( time taken to fall 10m minus time taken to fall 2.5m). In this time, C2 falls 10m. Now which formula do you think you should use?
 P: 4 Thank you for your reply! I think I was making things a bit too complicated. Would it work if I found the velocity of C1 after falling 2.5 m (I got -7.0 m/s, like I showed above), then found the time it took C1 to fall from 7.5 m to 0 m, using Formula 2: Formula 2: 0 = 7.5 + (-7.0)(t) + (1/2)(-9.81)(t)^2 When using the quadratic formula, I found time to equal 0.714 s. Then, I subbed this time into Formula 2 again, this time to find the initial velocity for C2: Formula 2: 0 = 10 + v(i)(0.714) + (1/2)(-9.81)(0.714)^2 I then found the velocity to be -9.1 m/s. So is that final answer of -9.1 m/s correct? Thank you very much to whoever replies!
HW Helper
P: 1,986

## 2 objects falling at diff. times; must find init. vel. of 2nd if land at same time

 Quote by unigal13 Would it work if I found the velocity of C1 after falling 2.5 m (I got -7.0 m/s, like I showed above), then found the time it took C1 to fall from 7.5 m to 0 m, using Formula 2:
You do not have to explicitly find the speed of C1 after it has fallen 2.5m. You just need to find the time C1 takes to fall 7.5m, as I have pointed out. Nevertheless, your method is OK. I have not checked the calculations.

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