A particle of mass $m$ is projected with a speed $u$ from the
ground at an angle $\theta=\frac{\pi}{3}$ w.r.t. horizontal (x-axis). When
it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity $u \hat{i}$. The horizontal distance covered by the combined mass before reaching the ground is:
Correct Option: 1
(1) Using principal of conservation of linear momentum for horizontal motion, we have
2 m v_{x}=m u+m u \cos 60^{\circ}
$v_{x}=\frac{3 u}{4}$
For vertical motion
$h=0+\frac{1}{2} g T^{2} \Rightarrow T=\sqrt{\frac{2 h}{g}}$
Let $R$ is the horizontal distance travelled by the body.
$R=v_{x} T+\frac{1}{2}(0)(T)^{2}$ (For horizontal motion)
$R=v_{x} T=\frac{3 u}{4} \times \sqrt{\frac{2 h}{g}}$
$\Rightarrow \quad R=\frac{3 \sqrt{3} u^{2}}{8 g}$
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