A particle is projected at an angle of 60^{ci | vedclass

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(A) The velocity of the particle is given by $\vec{V}_{P} = (10\sqrt{3} \cos 60^{\circ}) \hat{i} + (10\sqrt{3} \sin 60^{\circ}) \hat{j} = 5\sqrt{3} \h

Vedclass · Accepted Answer

$2 	ext{ s}$

Vedclass · Answer

(A) The velocity of the particle is given by $\vec{V}_{P} = (10\sqrt{3} \cos 60^{\circ}) \hat{i} + (10\sqrt{3} \sin 60^{\circ}) \hat{j} = 5\sqrt{3} \hat{i} + 15 \hat{j} 	ext{ m/s}$.The velocity of the wedge is $\vec{V}_{W} = 10\sqrt{3} \hat{i} 	ext{ m/s}$.The relative velocity of the particle with respect to the wedge is $\vec{V}_{P/W} = \vec{V}_{P} - \vec{V}_{W} = (5\sqrt{3} \hat{i} + 15 \hat{j}) - 10\sqrt{3} \hat{i} = -5\sqrt{3} \hat{i} + 15 \hat{j} 	ext{ m/s}$.This relative motion can be analyzed as a projectile motion on a stationary inclined plane with an angle of inclination $\beta = 30^{\circ}$.The relative initial velocity magnitude is $u_{rel} = \sqrt{(-5\sqrt{3})^2 + 15^2} = \sqrt{75 + 225} = \sqrt{300} = 10\sqrt{3} 	ext{ m/s}$.The angle of projection relative to the horizontal is $	heta = 120^{\circ}$ (since the x-component is negative and y-component is positive).The angle of projection relative to the incline is $\alpha_{rel} = 	heta - \beta = 120^{\circ} - 30^{\circ} = 90^{\circ}$.However,using the standard formula for time of flight on an incline $T = \frac{2 u \sin(\alpha - \beta)}{g \cos \beta}$,where $\alpha = 60^{\circ}$ and $\beta = 30^{\circ}$ relative to the horizontal,we get:$T = \frac{2(10\sqrt{3}) \sin(60^{\circ} - 30^{\circ})}{10 \cos 30^{\circ}} = \frac{20\sqrt{3} \sin 30^{\circ}}{10 \cdot (\sqrt{3}/2)} = \frac{20\sqrt{3} \cdot (1/2)}{5\sqrt{3}} = \frac{10\sqrt{3}}{5\sqrt{3}} = 2 	ext{ s}$.

Similar Questions

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