A particle of mass m is projected with a speed u from the ground at an angle $\theta = {\pi \over 3}$ 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 $u\widehat i$ . The horizontal distance covered by the combined mass before reaching the ground is:

Question

$2\sqrt 2 {{{u^2}} \over g}$

${{3\sqrt 3 } \over 8}{{{u^2}} \over g}$

${{3\sqrt 2 } \over 4}{{{u^2}} \over g}$

${5 \over 8}{{{u^2}} \over g}$

Solution

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