The human body can safely tolerate an acceleration 9.00 times that due to gravity. with what minimum radius of curvature may a pilot recover from a dive in pitching up the airplane if the airplanes speed is 770km/h at the end of the dive?
Please don't frickin' answer the question if you don't know how to solve it PROPERLY...it makes my miserable life even MORE tiring Jesus Christ!
v^2/r=9(9.8)
solve for r, change v to m/s first.
Uhm, forgive me but i really have a hard Time understanding this..so r=Squared root of 88,2/213.9 ?
kn kjn
To determine the minimum radius of curvature required for the pilot to recover from a dive while pitching up the airplane, we need to consider the acceleration experienced by the pilot and use the concept of centripetal acceleration.
The acceleration experienced by the pilot can be calculated using the formula:
acceleration = (v^2) / r
where,
v = velocity of the airplane at the end of the dive
r = radius of curvature needed to recover from the dive
In this case, the acceleration that a human body can tolerate is 9.00 times the acceleration due to gravity (9.8 m/s^2). Therefore, the acceleration experienced by the pilot should not exceed 9.00 times the acceleration due to gravity.
First, let's convert the airplane's speed from km/h to m/s:
770 km/h = (770 * 1000) / 3600 = 213.89 m/s
Next, let's calculate the acceleration experienced by the pilot:
acceleration = (213.89^2) / r
To ensure that the acceleration does not exceed 9.00 times the acceleration due to gravity, we can write the inequality as:
(213.89^2) / r <= 9.00 * 9.8
Now, let's solve for the minimum radius of curvature (r):
r >= (213.89^2) / (9.00 * 9.8)
Calculating this value, we get:
r >= 5169.27 m
Therefore, the minimum radius of curvature needed for the pilot to recover from the dive while pitching up the airplane is 5169.27 meters.