Question

While sitting on a tree branch 11m above the ground, you drop a chestnut. When the chestnut has fallen 2.7m , you throw a second chestnut straight down. What initial speed must you give the second chestnut if they are both to reach the ground at the same time?

I know I am supposed to use the formula
xf = xo + vot + 1/2at but I do not know how to start.

Answer 1

YOU MEAN:
xf = xo + vot + (1/2)a t^2

How long does it take the first chestnut to reach the ground?

11 = (1/2) g t^2
t = sqrt (22/9.81) = 1.5 seconds

so how long is the second one in the air?

t = 1.5 - sqrt(2*2.7/9.81)
= 1.5 - .74 = .76 seconds

so the second one has to go 11 meters down in .76 seconds
11 = Vi t + (1/2) a t^2
11 = Vi (.76) -4.9 (.76)^2

Vi = 18.2 m/sd

Answer 2

Nice explanation