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Hint: Assume throwing a ball vertically upwards and then find its time of flight in both with or without retardation due to the air using equations of motion. Then once we get both the time of flights, we can find the percentage increase or decrease in the time of flight by calculating the change.

Let us assume a ball is thrown vertically upwards with a velocity $u$, then at the maximum height, the velocity of the ball will become zero. Considering there is no retardation caused due to air then by using equation of motion $v = u + at$, where $v$ is the final velocity of ball at the maximum height, $u$ is the initial velocity, $a$ is the acceleration due to gravity and $t$ be the time of flight.

So, we can write, $0=u–gt\implies t = \dfrac{u}{g}$ ………. (i)

Now, considering the case, where there is also retardation due to air. So, effective acceleration will be the sum of acceleration due to gravity and due to air, that is one-tenth of gravity.

Therefore, effective acceleration $g’=g+\dfrac{g}{10}=\dfrac{11g}{10}$

Now, the time of flight, $T’=\dfrac{u}{\dfrac{11g}{10}}=\dfrac{10u}{11g}$ ………. (ii)

Thus, change in time of flight $=\delta T=T-T’=\dfrac{u}{g}-\dfrac{10u}{11g}=\dfrac{u}{11g}$

The percentage change in the time of flight $=\dfrac{\delta T}{T}\times 100\%=\dfrac{\dfrac{u}{11g}}{\dfrac{u}{g}}\times 100\%=\dfrac{100}{11} \% =9\%$

Hence, option a is the correct answer.

Note: In questions like this, it can also be asked to find the percentage increase or decrease in maximum heights attained, in that case also we will use the equation of motion on an object thrown vertically upward. We can also consider a projectile motion but, that will only make the solution lengthy.

__Complete step-by-step answer:__Let us assume a ball is thrown vertically upwards with a velocity $u$, then at the maximum height, the velocity of the ball will become zero. Considering there is no retardation caused due to air then by using equation of motion $v = u + at$, where $v$ is the final velocity of ball at the maximum height, $u$ is the initial velocity, $a$ is the acceleration due to gravity and $t$ be the time of flight.

So, we can write, $0=u–gt\implies t = \dfrac{u}{g}$ ………. (i)

Now, considering the case, where there is also retardation due to air. So, effective acceleration will be the sum of acceleration due to gravity and due to air, that is one-tenth of gravity.

Therefore, effective acceleration $g’=g+\dfrac{g}{10}=\dfrac{11g}{10}$

Now, the time of flight, $T’=\dfrac{u}{\dfrac{11g}{10}}=\dfrac{10u}{11g}$ ………. (ii)

Thus, change in time of flight $=\delta T=T-T’=\dfrac{u}{g}-\dfrac{10u}{11g}=\dfrac{u}{11g}$

The percentage change in the time of flight $=\dfrac{\delta T}{T}\times 100\%=\dfrac{\dfrac{u}{11g}}{\dfrac{u}{g}}\times 100\%=\dfrac{100}{11} \% =9\%$

Hence, option a is the correct answer.

Note: In questions like this, it can also be asked to find the percentage increase or decrease in maximum heights attained, in that case also we will use the equation of motion on an object thrown vertically upward. We can also consider a projectile motion but, that will only make the solution lengthy.

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