Answer:
v = 129.5 m / s
Explanation:
For this exercise we must use the conservation of mechanical energy.
Starting point. On the side of the first hill
Em₀ = U = m g h₁
Final point. On the other hill
Em_f = K + U = ½ m v² + mgh₂
where h2 is the latura of the ora hill
as they indicate that there is no friction, energy is conserved
Em₀ = Em_f
mgh₁ = ½ mv² + m gh₂
v² = 2 mg (h1-h2)
v = [tex]\sqrt{2m g( h_1 - h_2)}[/tex]
let's calculate
v = [tex]\sqrt{ 2 \ 599 \ ( 32-18)}[/tex]
v = 129.5 m / s
This speed is horizontal at the top of the hill
