Question

The friction of the air causes a vertical retardation equal to $10\%$ of the acceleration due to gravity. Take $g=10m/s^2$ . The maximum height and time to reach the maximum height will be decreased by

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

Answer 1

The correct option is A $9\%,9\%$

Vertical retardation$a=0.1g$
Neglecting air friction, maximum height $H=\frac{u^2}{2g},$ time $t=\frac{u}{g}$

Considering air friction, maximum height $H^{\prime }=\frac{u^2}{2(g+a)},$ time $t^{\prime }=\frac{u}{(g+a)}$
Percentage decrease in height $=\frac{H-H^{\prime }}{H}\times 100=\frac{\frac{u^2}{2g}-\frac{u^2}{2(g+a)}}{\frac{u^2}{2g}}\times 100=\frac{a}{g+a}\times 100=\frac{0.1g}{g+0.1g}\times 100=9$

Similarly, percentage decrease in time $=\frac{t-t^{\prime }}{t}\times 100=\frac{a}{g+a}\times 100=9$