Answer:
11.3 m/s
Explanation:
KE₁ = KE₂
½m₁v₁² = ½m₂v₂²
½ (2 kg) v² = ½ (4 kg) (8 m/s)²
v ≈ 11.3 m/s
The molarity remains the same so the ratio does not change
<u>Answer:</u>
The height of ramp = 124.694 m
<u>Explanation:</u>
Using second equation of motion,

From the question,
u = 31 m/s; s = 156.3 m, a=0
substituting values

t = 
= 5.042 s
Similary, for the case of landing
t = 5.042 s; initial velocity, u =0
acceleration = acceleration due to gravity, g = 9.81 
Substituting in 

h = 124.694 m
So height of ramp = 124.694 m