At what height must the plane start a quarter turn to pull out of the dive?

A jet is diving vertically downward at 1390 km/h. If the pilot can stand a maximum acceleration of 5g (i.e., five times the acceleration of gravity on Earth's surface) before losing consciousness, at what height must the plane start a quarter turn to pull out of the dive? Assume the speed remains constant.|||If the jet's speed remains 1390 km/hr, and the pilot pulls up in a circular arc, the acceleration the pilot feels is





TotalAccel = g + a





where g is gravity and a is the angular acceleration caused by the turn.





Solving for a, we find the radius of the turn





5*g = g + a


4*g = a


4*g = v^2 / R


4*(9.81 m/s^2) = (386.1 m/s)^2 / R


(39.23 m/s^2) = (149100 m^2/s^2) / R


R = (149100 m^2/s^2) / (39.23 m/s^2)


R = 3800 m





So, the pilot must start pulling out of the dive at least 3800 meters (3.8 kilometers) above the ground to avoid both blacking out and colliding with the ground..|||Let's assume that pilot is trying to make shortes possible quarter turn. He should do it by moving on a circle with minimal possible radius. When plane is moving by circle everything in the plane experiences centripetal acceleration v^2/R. According to your problem statement this value should be no more than 5g, so v^2/R%26lt;5g, R%26gt;v^2/5g. Convert your speed to m/s by dividing 1390/3.6. The resut is in meters.





This solution is good for person who just started studying circular motion.


There are additional factors to consider. Acceleration 5g that the pilot experiences at the bottom of the loop means that one g is provided by gravity and 4g is centripetal acceleration, so in fact the radius at the bottom of the loop should be larger, so that v^2/R%26lt;4g, R%26gt;v^2/4g.





I do not know your level so I can not say for sure which answer your textbook would prefer:)





For AP Physics BC course you could take into consideration that the radius can vary from R%26gt;v^2/5g at the beginning of the turn to R%26gt;v^2/4g at the end of it.