Question

Two particles are projected vertically upwards from the surface of the earth with velocities $v_{1}=\sqrt{\frac{2 g R}{3}}$ and $v_{2}=\sqrt{\frac{4 g R}{3}}$ respectively. If the maximum heights attained by the two particles are $h_{1}$ and $h_{2}$ respectively, then calculate the ratio $\frac{h_{2}}{h_{1}}$ .

Answer 1

Loss of $KE=$ Gain in $PE$
$\Rightarrow \frac{1}{2} m v^{2}=m g h\left(\frac{ R }{ h + R }\right)$
CASE I.
$\Rightarrow \frac{1}{2}m\left(\frac{2 gR}{3}\right)=\left(mgh\right)_{1}\left(\frac{R}{h_{1} + R}\right)$ ...(i)
$\Rightarrow \frac{1}{2}m\left(\frac{4 g R}{3}\right)=\left(mgh\right)_{2}\left(\frac{R}{h_{2} + R}\right)$ ...(ii)
Solving (i)
$\Rightarrow \frac{1}{3}=\frac{h_{1}}{h_{1} + R}$
$\Rightarrow h_{1}=\frac{R}{2}$ ...(iii)
Solving (ii)
$\Rightarrow \frac{2}{3}=\frac{h_{2}}{h_{2} + R}$ ...(iv)
$\Rightarrow h_{2}=2R$
From (iii) and (iv)
$\Rightarrow \frac{h_{1}}{h_{2}}=\frac{R / 2}{2 R}$
$\therefore \, \, \frac{h_{1}}{h_{2}}=\frac{1}{4}=0.25$