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 $g= 10 \frac{m}{s^2}$
My attempt: Only the vertical component of the projectile will be affected by changes in $g$ so by doing the time of flight analysis for a particle thrown up with velocity equal to the vertical component of the velocity of projectile, the analysis will become easier.
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 $1 \frac{m}{s^2}$. So, when the ball is thrown up, net acceleration equals $10+1$=$11 \frac{m}{s^2}$ 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 $10-1$=$9\frac{m}{s^2}$ in the downward direction.
Time of flight=$\frac{u}{9}+\frac{u}{11}$ where u is the velocity of projectile in the vertical direction.
In the absence of air resistance, time of flight=$\frac{2u}{g}$=$\frac{u}{5}$
I think this analysis is wrong because it is not leading me to the answer. Where am I wrong?
Best Answer
Hint: I think the issue here is that the motion here is not symmetric. If it starts out with a speed $u$, it is not necessary that it will have the same speed when it reaches the bottom - because the acceleration is not the same in both the cases.
In the case without air resistance, it is valid to write $t=\frac{u}{g}+\frac{u}{g}$, because the particle goes from $u$ to $0$ and then $0$ to $u$, which doesn't happen in this case.
Try studying the problem by taking into account the distance the projectile travels - because the distance it travels up is always the distance it'll travel down.