Question

Two particles A and B are projected from a point O on a horizontal plane, with velocities 20m/s and $20\sqrt{2m/s}$ respectively at angles ${30}^{\circ }$ and ${45}^{\circ }$ with the horizontal respectively in the same plane. If the maximum separation is $x\times 10\sqrt{1^2+(\sqrt{3}-2{)}^2}$. Find x. Assume that the two particles get stuck to the ground when they fall over it.___

Answer 1

$T_A=\frac{2\times 20\times \frac{1}{2}}{10}=2sec$
$T_B=\frac{2\times 20\sqrt{2}\times \frac{1}{\sqrt{2}}}{10}=4sec$
$S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2=[(20\frac{\sqrt{3}}{2}\hat{i}+20\frac{1}{2}\hat{j})t-(20\sqrt{2}\frac{1}{\sqrt{2}}\hat{i}+20\sqrt{2}\frac{1}{2}\hat{j})t]$
$=(10\sqrt{3}\hat{i}+10\hat{j}-20\hat{j})t=\left[10\right(\sqrt{3}-2)\hat{i}-10\hat{j}]t$
$S_{max}=20(\sqrt{3}-2)\hat{i}-20\hat{j}$, at t=2
$=20\sqrt{1^2+(\sqrt{3}-2{)}^2}\approx 40m$