The friction of air causes a vertical retardation equal to 10% of the acceleration due to gravity (take = $g = 10 m / s^{2}$ ). The maximum height will be decreased by

11%

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12%

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20%

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Solution

The correct option is B 9%
Maximum height, $h_{1} = \frac{u^{2}}{10}$

With friction :
Maximum height, $h_{2} = \frac{u^{2}}{g + 10 % g} = \frac{u^{2}}{11}$

% change = $\frac{h_{2} - h_{1}}{h_{1}} = \frac{(\frac{u^{2}}{11} - \frac{u^{2}}{10})}{(\frac{u^{2}}{10})} = - 9 %$

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The friction of air causes a vertical retardation equal to 10% of the acceleration due to gravity (take = g=10m/s2). The maximum height will be decreased by

The correct option is B 9% Maximum height, h1=u210 With friction : Maximum height, h2=u2g+10%g=u211 % change = h2−h1h1=(u211−u210)(u210)=–9%

The friction of air causes a vertical retardation equal to 10% of the acceleration due to gravity (take = $g = 10 m / s^{2}$ ). The maximum height will be decreased by

The correct option is B 9%
Maximum height, $h_{1} = \frac{u^{2}}{10}$

With friction :
Maximum height, $h_{2} = \frac{u^{2}}{g + 10 % g} = \frac{u^{2}}{11}$

% change = $\frac{h_{2} - h_{1}}{h_{1}} = \frac{(\frac{u^{2}}{11} - \frac{u^{2}}{10})}{(\frac{u^{2}}{10})} = - 9 %$