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<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
Given that,
Initial velocity , Vi = 0
Final velocity , Vf = 40 m/s
Acceleration due to gravity , a = 9.81 m/s²
Distance can be calculated as,
2as = Vf² - Vi²
2 * 9.81 *s = 40² - 0²
s = 81.55 m
For half height, that is, s = 40.77m
Vf= ??
2as = Vf² - Vi²
2 * 9.81 * 40.77 = Vf² - 0²
Vf² = 800
Vf = 28.28 m/s
Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.
Answer:
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