Question

A particle is thrown upwards with speed $v_0$. Find its speed when it reaches $2/3^{rd}$ of the maximum height.

• A. $\frac{v_0}{\sqrt{2}}$
• B. $\frac{2v_0}{3}$
• C. $\frac{v_0}{\sqrt{3}}$
• D. $\frac{v_0}{3}$

Answer 1

The correct option is C $\frac{v_0}{\sqrt{3}}$

Maximum height of the particle is given as
$H_{max}=\frac{u^2}{2g}=\frac{v_0^2}{2g}$
So, the speed of the particle at the height $(\frac{2}{3}\times H_{max}=\frac{v_0^2}{3g})$ is given as
$v^2-u^2=2as$
$\Rightarrow v^2-v_0^2=2\times (-g)\times \frac{v_0^2}{3g}$
$\Rightarrow v^2=v_0^2-\frac{2v_0^2}{3}=\frac{v_0^2}{3}$
$\Rightarrow v=\frac{v_0}{\sqrt{3}}$