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(C) Initial velocity $u = 50 \ m/s$,angle $	heta = 30^{\circ}$,$g = 10 \ m/s^2$.Horizontal component of velocity $u_x = u \cos 	heta = 50 \cos 30^{\

Vedclass · Accepted Answer

$86.6$

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(C) Initial velocity $u = 50 \ m/s$,angle $	heta = 30^{\circ}$,$g = 10 \ m/s^2$.Horizontal component of velocity $u_x = u \cos 	heta = 50 \cos 30^{\circ} = 50 	imes \frac{\sqrt{3}}{2} = 25\sqrt{3} \ m/s$.Vertical component of velocity $u_y = u \sin 	heta = 50 \sin 30^{\circ} = 50 	imes 0.5 = 25 \ m/s$.Total range $R = \frac{u^2 \sin 2	heta}{g} = \frac{50^2 \sin 60^{\circ}}{10} = \frac{2500 	imes \sqrt{3}}{20} = 125\sqrt{3} \ m \approx 216.5 \ m$.Horizontal distance covered in $t = 3 \ s$ is $x = u_x 	imes t = 25\sqrt{3} 	imes 3 = 75\sqrt{3} \ m \approx 129.9 \ m$.The distance beyond the wall where the stone strikes the ground is $R - x = 125\sqrt{3} - 75\sqrt{3} = 50\sqrt{3} \ m$.$50 	imes 1.732 = 86.6 \ m$.

$A$ stone projected from the ground with a velocity $50 \ m/s$ at an angle of $30^{\circ}$ with the horizontal crosses a wall after a time of $3 \ s$. Then the horizontal distance beyond the wall that the stone strikes the ground is (acceleration due to gravity $g = 10 \ m/s^2$) (in $m$)

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