If retardation produced by air resistances to projectile is one-tenth of acceleration due to gravity, the time to reach miximum height approximately-

Increase by

9

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Decrease by

9

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Increase by

11

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Decrease by

11

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Solution

The correct option is B Decrease by $9$ %
time taken to reach the top
$v = u + a t$
$0 = u s i n θ - (\frac{11 g}{10}) t$

$t = \frac{10 u s i n θ}{11 g}$
the decrement is $\frac{\frac{u s i n θ}{g} - \frac{10 u s i n θ}{11 g}}{\frac{u s i n θ}{g}} \times 100 = \frac{1}{11} \times 100 = 9$ %

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If retardation produced by air resistances to projectile is one-tenth of acceleration due to gravity, the time to reach miximum height approximately-

The correct option is B Decrease by 9%time taken to reach the topv=u+at0=usinθ−(11g10)tt=10usinθ11gthe decrement is usinθg−10usinθ11gusinθg×100=111×100=9%

The correct option is B Decrease by $9$ %
time taken to reach the top
$v = u + a t$
$0 = u s i n θ - (\frac{11 g}{10}) t$

$t = \frac{10 u s i n θ}{11 g}$
the decrement is $\frac{\frac{u s i n θ}{g} - \frac{10 u s i n θ}{11 g}}{\frac{u s i n θ}{g}} \times 100 = \frac{1}{11} \times 100 = 9$ %