Question

A jet plane comes in for a downward dive as shown in the figure below.
The bottom part of the path is a quarter circle having a radius of curvature of 350m . According to medical tests, pilots lose consciousness at an acceleration of 5.50g. At what speed (in m/s) will the pilot black out for this dive?

Answer 1

The solution would be like this for this specific problem:

5.5 g = g + v^2/r 
4.5 g = v^2/r 
v^2 = 4.5 g * r 
v = sqrt ( 4.5 *9.81m/s^2 * 350 m) 
v = 124 m/s

So the pilot will black out for this dive at 124 m/s. I am hoping that these answers have satisfied your query and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

Answer 2

As we know that centripetal acceleration during circular motion is given by the formula

$a = \frac{v^2}{R}$

as we know that maximum value of acceleration is given as

$a = 5.50 g$

here we know that

$g = 9.81 m/s^2$

so here we will have

$a = 5.50 (9.81 m/s^2)$

$a = 54 m/s^2$

now we will have from above first equation

$54 = \frac{v^2}{R}$

$54 = \frac{v^2}{350}$

$v = 137.4 m/s$

so maximum possible speed will be 137.4 m/s