Question

A stone with weight $w$ is thrown vertically upward into the air from ground level with initial speed $v_{0}$. If a constant force $f$ due to air drag acts on the stone throughout its flight. The maximum height attained by the stone is

• A. $h=\frac{v_{0}^{2}}{2 g\left(1+\frac{f}{w}\right)}$
• B. $h=\frac{v_{0}^{2}}{2 g\left(1-\frac{f}{w}\right)}$
• C. $h=\frac{v_{0}^{2}}{ g\left(1-\frac{w}{f}\right)}$
• D. $h=\frac{v_{0}^{2}}{2 g\left(1-\frac{w}{f}\right)}$

Answer 1

Answer: (A)

Using work energy theorem
$W_{f}+W_{g}=\Delta K$
$-f . h-W h=0-\frac{1}{2} m v_{0}^{2}$
$h=\frac{v_{0}^{2}}{2 g\left(1+\frac{f}{w}\right)}$