Question

A stone is thrown vertically upward with an initial velocity $V_0$. The distance traveled in time $4v_0/3g$ is

• A. $\frac{2v_0^2}{g}$
• B. $\frac{v_0^2}{2g}$
• C. $\frac{4v_0^2}{9g}$
• D. $\frac{5v_0^2}{9g}$

Answer 1

Answer: (C)

$\frac{4v_0^2}{9g}$

Lets consider,

$v_0=$ initial velocity of the stone

$g=$ acceleration due to gravity

$t=\frac{4v_0}{3g}$

2nd equation of motion,

$s=ut-\frac{1}{2}g{t}^2$

$s=v_0\frac{4v_0}{3g}-\frac{1}{2}g(\frac{4v_0}{3g}{)}^2$

$s=\frac{4v_0^2}{3g}-\frac{8v_0^2}{9g}$

$s=\frac{4v_0^2}{9g}$

The correct option is C.