Question

the maximum normal force a pilot can withstand is about eight times his weight. What is the maximum radius of curvature that a jet planes pilot, pulling out of a vertical dive

Answer 1

Complete Question

the maximum force a pilot can stand is about seven times his weight. what is the minimum radius of curvature that a jet plane's pilot, pulling out of a vertical dive, can tolerate at a speed of 250m/s?

Answer:

The value is    $r = \frac{250^2 }{6 * 9.8 }$

Explanation:

From the question we are told that

 The  weight of the pilot is   $W = mg$

 The maximum force a pilot can withstand is  $F_{max} = 7 W = 7 (mg)$

 The speed is  $v = 250 \ m/s$  

Generally the centripetal force acting on the pilot is equal to the net force acting on the pilot i.e

      $F_c = F_{max} - mg$

Here N  is the normal force acting on the pilot

Now

      $F_c = \frac{m v^2 }{r}$

So

      $\frac{m v^2 }{r} = 7(mg) - mg$  

=>  $r = \frac{v^2 }{6g}$

=>  $r = \frac{250^2 }{6 * 9.8 }$

=>  $r = 1063 \ m$