Three charged particles A, B and C are thrown at the same time same point from the top a very high building with velocities (in m/s) mentioned \left(\overset{\rightarrow}{\text{V}}\right)_{\text{A}} = 3 \hat{j} - \hat{i} ; \left(\overset{\rightarrow}{\text{V}}\right)_{\text{B}} = 3 \hat{j} ; \left(\overset{\rightarrow}{\text{V}}\right)_{\text{C}} = 6 \hat{j} + 2 \hat{i} All the particles are of the same mass. Find the maximum height attained by the centre of mass of system of three particles (in cm).

Engineering

Physics

Newtons Second Law of Motion

Collision

Relative Motion

Question

Three charged particles A, B and C are thrown at the same time same point from the top a very high building with velocities (in m/s) mentioned
${\vec{V}}_{A} = 3 \hat{j} - \hat{i}; {\vec{V}}_{B} = 3 \hat{j}; {\vec{V}}_{C} = 6 \hat{j} + 2 \hat{i}$
All the particles are of the same mass.

Find the maximum height attained by the centre of mass of system of three particles (in cm).

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Velocities mentioned are

${\vec{V}}_{A} = 3 \hat{j} - \hat{i}; {\vec{V}}_{B} = 3 \hat{j}; {\vec{V}}_{C} = 6 \hat{j} + 2 \hat{i}$

$V_{cm} \frac{- 1 \hat{i} + 3 \hat{j} + 3 \hat{j} + 6 \hat{j} + 2 \hat{i}}{3}$

$V_{cm} = 4 \hat{j} m / s + \frac{1 \hat{i}}{3}$

$H = \frac{V_{cm}^{2}}{2 g} = \frac{4^{2}}{2 \times 10} = \frac{816}{20} = 0 .8 m$