Answer:
u = 14 m / s
Explanation:
For this exercise let's use conservation of energy
starting point. On the floor just when u speeding out
Em₀ = K = ½ m v²
final point. When on top of the building, no speed
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ m v² = m g h
v = u
u = [tex]\sqrt{2 g h}[/tex]
u= [tex]\sqrt{2 \ 9.8 \ 10}[/tex]
u = 14 m / s