The friction of the air causes vertical retardation equal to one tenth of the acceleration due to gravity (Take g = $10 m s^{- 2}$ ). The time of flight will be decreased by

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Solution

The correct option is C

$T = \frac{2 u s i n θ}{g} ∴ \frac{T_{1}}{T_{2}} = \frac{g_{2}}{g_{1}} = \frac{g + \frac{g}{10}}{g} = \frac{11}{10}$

Fractional decrease in time of flight $= \frac{T_{1} - T_{2}}{T_{1}} = \frac{1}{11}$

Percentage decrease $= 9 %$

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

The correct option is C T=2u sin θg ∴T1T2=g2g1=g+g10g=1110 Fractional decrease in time of flight =T1−T2T1=111 Percentage decrease =9%

The friction of the air causes vertical retardation equal to one tenth of the acceleration due to gravity (Take g = $10 m s^{- 2}$ ). The time of flight will be decreased by

The correct option is C

$T = \frac{2 u s i n θ}{g} ∴ \frac{T_{1}}{T_{2}} = \frac{g_{2}}{g_{1}} = \frac{g + \frac{g}{10}}{g} = \frac{11}{10}$

Fractional decrease in time of flight $= \frac{T_{1} - T_{2}}{T_{1}} = \frac{1}{11}$

Percentage decrease $= 9 %$