Terminal speed=136.7 m/s
the formula for the terminal speed is given by :
V= 
m= mass=75 kg
a= area=0.2 x 0.4=0.08 m²
ρ= density of air=1.23 kg/m³
C= drag constant =0.8
so V= 
v= 136.7 m/s
Answer:
The time taken by the airplane to take off, t = 11.46 s
Explanation:
Given data,
The initial velocity of the airplane, u = 24 m/s
The acceleration of the plane, a = 8 m/s
The distance covered until take off, d = 800 m
Using the III equation of motion,
v² = u² +2as
= 24² + 2 x 8 x 800
= 13376
v = 115.65 m/s
Using the first equation of motion,
v = u + at
t = (v-u) / a
= (115.65 - 24) / 8
= 11.46 s
Hence, the time taken by the airplane to take off, t = 11.46 s
A Newton is a derived unit.
Hope that helped:)
