Particle is projected with speed $5 \sqrt{2} m/s$ at an angle of projection $45^{\circ}$ from horizontal. At maximum height it breaks in two identical particle, velocity of one particle is zero, find out distance (in $m$ ) of second particle from point of projection when it hits ground. Take $g = 10 {m/s}^{2}$

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Solution

Using position of COM
$r_{C M} = \frac{m_{1} r_{1} + m_{2} r_{2}}{m_{1} + m_{2}} \Rightarrow R = \frac{\frac{m}{2} \times \frac{R}{2} + \frac{m}{2} \times x}{m}$
$\Rightarrow R = \frac{R}{4} + \frac{x}{2}$
$\Rightarrow \frac{3 R}{4} = \frac{x}{2}$
$\Rightarrow x = \frac{3 R}{2} = \frac{3}{2} \times \frac{u^{2} sin 2 θ}{g} = \frac{3}{2} \times \frac{50 \times 1}{10} = 7.5 m$

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Particle is projected with speed 5√2 m/s at an angle of projection 45∘ from horizontal. At maximum height it breaks in two identical particle, velocity of one particle is zero, find out distance (in m) of second particle from point of projection when it hits ground. Take g=10 m/s2

Using position of COM rCM=m1r1+m2r2m1+m2⇒R=m2×R2+m2×xm ⇒ R=R4+x2 ⇒3R4=x2 ⇒ x=3R2=32×u2sin2θg=32×50×110=7.5 m