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Sagot :
Answer:
[tex]63.38\ \text{mph}[/tex]
Explanation:
L = Lift force
[tex]\rho[/tex] = Density of air
A = Surface area
v = Velocity
[tex]v_1[/tex] = 45 mph
[tex]\rho_1=1.23\ \text{kg/m}^3[/tex]
[tex]\rho_2=0.62\ \text{kg/m}^3[/tex]
Coefficient of lift is given by
[tex]CL=\dfrac{2L}{\rho v^2A}\\\Rightarrow \rho=\dfrac{2L}{CL v^2A}[/tex]
So
[tex]\rho\propto \dfrac{1}{v^2}[/tex]
[tex]\dfrac{\rho_1}{\rho_2}=\dfrac{v_2^2}{v_1^2}\\\Rightarrow v_2=\sqrt{\dfrac{\rho_1}{\rho_2}}\times v_1\\\Rightarrow v_2=\sqrt{\dfrac{1.23}{0.62}}\times 45\\\Rightarrow v_2=63.38\ \text{mph}[/tex]
The velocity at the required altitude should be [tex]63.38\ \text{mph}[/tex] to maintain the same lift.
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