A golf ball is thrown straight up from the edge of the roof

A golf ball is thrown straight up from the edge of the roof of a building. A second golf ball is dropped from the roof a time of 1.05 s later. You may ignore air resistance.

If the height of the building is 20.7 m , what must the initial speed be of the first ball if both are to hit the ground at the same time?

Solution

y= -20.0m, Vo2 = 0, g = -9.8 m/s^2

first I solved for the time of the 2nd ball to hit the ground:

y = u.t + (1/2)(g)(t^2)
-20.7m = 0 + (1/2)(-9.8 m/s^2)(t^2)
-20.7m = (-4.9m/s^2)(t^2)
4.224 = t^2

t = 2.055 s

time for the 1st ball to hit the ground = 2.055s + 1.05s = 3.105s

Now I solve for initial speed of first ball:

-20.7m = u (3.105s) - (4.9 m/s^2) (3.105s)^2
-20.7m + 47.24m = u (3.105s)

u = 8.547 m/s