The friction of the air causes vertical retardation equal to one tenth of the acceleration due to gravity (Take g = 10 ms⁻²). The time of flight will be decreased by

The correct option is 9%.

Let us analyze the motion of the body with and without air resistance to find the percentage decrease in the time of flight.

Case 1: Without Air Resistance
When a body is projected vertically upwards with an initial velocity $u$ under gravity alone, the acceleration acting on it is the acceleration due to gravity, $g$ (acting downwards).
The total time of flight $T$ is given by the formula:

$T=\frac{2u}{g}$

Case 2: With Air Resistance
Air friction acts opposite to the direction of motion, causing a retardation. The problem specifies that the vertical retardation due to air friction is equal to one-tenth of the acceleration due to gravity:
$a_r=0.1g$
During the motion, the effective deceleration $g^{\text{'}}$ resisting the upward motion is the sum of gravity and air retardation:

$g^{\text{'}}=g+0.1g=1.1g$

Using this effective deceleration, the new time of flight $T^{\text{'}}$ can be approximated as:

$T^{\text{'}}=\frac{2u}{g^{\text{'}}}=\frac{2u}{1.1g}$

Calculating the Percentage Decrease:
The fractional decrease in the time of flight is:

$\frac{T-T^{\text{'}}}{T}=1-\frac{T^{\text{'}}}{T}$

Substituting the values of $T$ and $T^{\text{'}}$:

$1-\frac{1}{1.1}=1-\frac{10}{11}=\frac{1}{11}$

Converting this to a percentage:

$\text{Percentage Decrease}=\frac{1}{11}\times 100\%\approx 9.09\%$

Rounding to the nearest integer option gives 9%.