Question

The friction of air causes vertical retardation equal to one tenth of acceleration due to gravity (g = 10 m/s 2 ). By how much percent will be the time of flight decreased ?

Answer 1

Time of flight T=2u sinθ/gθ T1/T2 =g2/g1 = (g+g/10) /g =11/10 Fractional decrease in time of flight=(T1−T2) /T1 =[ (T1/T2)−1 ] / (T1/T2) = [ (11/10−1) ] / 11/10= 1/11 So the % decrease in flight time will be =9%

Answer 2

The decreased time of flight percent is 9%

Given:

Friction of air = Retardation in vertical direction

Acceleration due to gravity = 1/10 th

Solution:

The time of flight is given by the formula:

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

Therefore, two time of flight be,

$\frac{T 1}{T 2}=\frac{g 2}{g 1}$

$\frac{T 1}{T 2}=\frac{g+\frac{g}{10}}{g}$

$\frac{T 1}{T 2}=\frac{11}{10}$

Thereby, the decrease in time of flight be,

$\frac{T 1-T 2}{T 1}=1-\frac{T 2}{T 1}$

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

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

$\frac{T 1-T 2}{T 1}=\frac{1}{11}$

Thereby, in percentage it will come out to be 9 percent decrease in time of flight.