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. $h=\frac{v_0^2}{2g(1+\frac{f}{W})}$
• B. $h=\frac{v_0^2}{2g(1-\frac{f}{W})}$
• C. $h=\frac{v_0^2}{3g(1+\frac{f}{W})}$
• D. $h=\frac{v_0^2}{2g(1+\frac{2f}{W})}$

Answer 1

The correct option is A $h=\frac{v_0^2}{2g(1+\frac{f}{W})}$

Using work energy theorem, $W_f+W_g=\Delta K$

Where $W_f$ is the work done by the air drag and $W_g$ is the work done by gravity

$-f.h-Wh=0-\frac{1}{2}m{v}_0^2$
$fh+Wh=\frac{W{v}_0^2}{2g}$

$h=\frac{v_0^2}{2g(1+\frac{f}{W})}$

Hence option A is the correct answer