Answer:
The height of ramp = 124.694 m
Explanation:
Using second equation of motion,
[tex]s = ut + \frac{1}{2}at^2[/tex]
From the question,
u = 31 m/s; s = 156.3 m, a=0
substituting values
[tex]156.3 = 31\times t + 0[/tex]
t = [tex]\frac{156.3}{31 }[/tex]
= 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 [tex]m/s^2[/tex]
Substituting in [tex]h = ut + \frac{1}{2}gt^2[/tex]
[tex]h = 0 + \frac{1}{2} \times 9.81 \times (5.042)^2[/tex]
h = 124.694 m
So height of ramp = 124.694 m
