Question

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

• A. $8\%$
• B. $9\%$
• C. $10\%$
• D. $11\%$

Answer 1

The correct option is B $9\%$

The correct option is B.

Given,

Retardation $=10\%$

If there is no friction then,

By the equation of the motion,

$v^2=u^2+2gh$

$v=0$

$h=\frac{u^2}{2g}$

Now,

When there is air resistance $10\%$ of the gravity.

$g_{eff}=g+0.1g=1.1g$

The maximum height will be,

$h_1=\frac{u^2}{2g_{ef}}$

Therefore, the difference is $h_1-h$

$u^{2}2g_{ef}-\frac{u^2}{2g}$

So,

Change in height is:

$\frac{u^2}{2g}-\frac{u^2}{2g_{ef}}$

$=\frac{u^2}{2g}-\frac{u^2}{2g}(1-\frac{11}{10})$

Thus the percentage change in height is:

$=\frac{\frac{u^2}{2g}(1-\frac{11}{10})}{\frac{u^2}{2g}}\times 100$

$=\frac{100}{11}\approx 9\%$