Answer: The minimum acceleration for the air plane is 2.269m/s2.
Explanation: To solve such problem the equation of motion are applicable.
The initial velocity is 0 since the airplane was initially standing. We are going to use this equation
V^2=U^2+2as
33^2=0+2a (240)
a= 2.269m/s2
Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s
To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s
Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m
This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.
But anyways, good luck!
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