To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,


Here,
m = mass
k = Spring constant
x = Displacement
v = Velocity
Rearranging to find the velocity,



Our values are,



Replacing our values we have,


Therefore the velocity is 
<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters
</span>
Answer:
λ = 5.65m
Explanation:
The Path Difference Condition is given as:
δ=
;
where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.
m = no of openings which is 2
∴δ= 
n is the index of refraction of the medium in which the wave is traveling
To find δ we have;
δ= 
δ= 
δ= 
δ= 
δ= 
δ= 
δ= 82.15 -73.68
δ= 8.47
Again remember; to calculate the wavelength of the ocean waves; we have:
δ= 
δ= 8.47
8.47 = 
λ = 
λ = 5.65m
Faster and higher I believe.