Air resistnce in projectile motion decreasing time of flight

[Elena14]

If the air resistance causes a vertical retardation of 10 percent of value of acceleration due to gravity, then the time of flight of a projectile will be decreased by nearly? Take g=10m/s^2

My attempt: Only the vertical component of the projectile will be affected by changes in g

I understand that air resistance is like friction; it acts in the direction opposite to the motion. So, when the particle is thrown up, the air resistance together with gravity will act in downward direction. According to the question, value of retardation due to air resistance is 1m/s^2. So, when the ball is thrown up, net acceleration equals 10+1=11m/s^2 in the downward direction.
When the particle is coming down after reaching its maximum height, air resistance should oppose its motion and act in vertically upward direction. Therefore, net acceleration must be 10-1=9m/s^2 in the downward direction.

Time of flight =u/9 + u/11 where u is the velocity of projectile in the vertical direction.

In the absence of air resistance, time of flight=2u/g=u/5

I think this analysis is wrong because it is not leading me to the answer. Where am I wrong?

 

[jbriggs444]

According to the question, value of retardation due to air resistance is 1m/s^2. So, when the ball is thrown up, net acceleration equals 10+1=11m/s^2 in the downward direction.

If "u" denotes initial upward velocity then indeed time to max height would then be given by u/11. I agree with that part of your answer.

When the particle is coming down after reaching its maximum height, air resistance should oppose its motion and act in vertically upward direction. Therefore, net acceleration must be 10-1=9m/s^2 in the downward direction.

I agree with this. But then you come up with a result of u/9 for the time to fall. That would be the time needed to regain a velocity of -u, not the time required to fall from a height that you have not calculated.

By the time the object accelerating at 9 meters per second per second downward had regained its initial speed (now in the downward direction), it would be well below its starting position.