What is the speed of the ball as it passes the top of the window?

Question

A ball is thrown vertically downward from the top of a 38.5-m-tall building. The ball passes the top of a window that is 13.9 m above the ground 2.00 s after being thrown. What is the speed of the ball as it passes the top of the window?

Answer 1

s = v0t + (1/2)gt^2

t = 2 s
s = 38.5 - 13.9 = 24.6 m
g = 9.8 m/s^2

24.6 = v0(2) + (1/2)9.8(2^2)
24.6 = 2v0 + 19.6
5 = 2v0
v0 = 2.5 m/s

v = v0 + at
v = 2.5 + 9.8(2)
v = 22.1 m/s

Answer 2

s = [ 38.5 - 13.9] =24.6 m

v-u = gt = 9.8 x 2 = 19.6m
[v +u] /2 * t = 24.6 m Since s = average velocity x time.
[v +u] = 24.6 m/s since t = 2 second

Adding [v +u] and [v-u]
2v = 44.2m/s
v =22.1m/s.

[u = 2.5m/s] not given.and not asked