Question

A conventional airplane of mass 5000 kg is doing a level turn of radius 1000 m at a constant speed of 100 m/s at sea level.
Taking the acceleration due to gravity as 10 m/s\(^2\), the bank angle of the airplane is _________ degrees.

Hint:

The bank angle in level turns is determined by the ratio of the square of the aircraft's velocity to the product of the gravitational acceleration and the turn radius. Use this formula to quickly calculate the bank angle.

Answer 1

Correct Answer: 44

In a level turn, the aircraft must generate a centripetal force to stay on its circular path. The centripetal force is provided by the horizontal component of the lift. The equation for the bank angle \( \theta \) is given by: \[ \tan \theta = \frac{v^2}{g \times r}, \] where \( v = 100 \, \text{m/s} \) is the velocity, \( g = 10 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( r = 1000 \, \text{m} \) is the radius of the turn. Substituting the values: \[ \tan \theta = \frac{(100)^2}{10 \times 1000} = \frac{10000}{10000} = 1. \] Thus, \( \theta = \tan^{-1}(1) = 45^\circ \). So, the bank angle is: \[ \boxed{45^\circ}. \]