Question

Your boss asks you to design a drone that begins its flight near the surface and rises to 9600 m. At the surface it will fly through air having a density of 1.23 kg per cubic meter and at its highest altitude the air density will become 0.62 kg per cubic meter. If the flight velocity near sea level is 45 mph, then how fast will in need to go at its highest altitude to maintain the same lift. Assume the coefficient of lift remains constant.

Answer 1

`63.38\ \text{mph}`

Step-by-step explanation:

L = Lift force

`\rho` = Density of air

A = Surface area

v = Velocity

`v_1` = 45 mph

`\rho_1=1.23\ \text{kg/m}^3`

`\rho_2=0.62\ \text{kg/m}^3`

Coefficient of lift is given by

`CL=(2L)/(\rho v^2A)\\\Rightarrow \rho=(2L)/(CL v^2A)`

So

`\rho\propto (1)/(v^2)`

`(\rho_1)/(\rho_2)=(v_2^2)/(v_1^2)\\\Rightarrow v_2=\sqrt{(\rho_1)/(\rho_2)}* v_1\\\Rightarrow v_2=\sqrt{(1.23)/(0.62)}* 45\\\Rightarrow v_2=63.38\ \text{mph}`

The velocity at the required altitude should be
`63.38\ \text{mph}` to maintain the same lift.