A body of mass 1,kg is projected with velocit | vedclass

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(D) The initial horizontal velocity is $u_x = u \cos 	heta = 50 \cos 30^{\circ} = 50 	imes \frac{\sqrt{3}}{2} = 25 \sqrt{3}\,m/s$.The initial vertic

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$250 \sqrt{3}\,m$

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(D) The initial horizontal velocity is $u_x = u \cos 	heta = 50 \cos 30^{\circ} = 50 	imes \frac{\sqrt{3}}{2} = 25 \sqrt{3}\,m/s$.The initial vertical velocity is $u_y = u \sin 	heta = 50 \sin 30^{\circ} = 50 	imes \frac{1}{2} = 25\,m/s$.The time to reach the highest point is $t_1 = \frac{u_y}{g} = \frac{25}{10} = 2.5\,s$.The horizontal distance covered during this time is $x_1 = u_x t_1 = (25 \sqrt{3})(2.5) = 62.5 \sqrt{3}\,m$.At the highest point,the vertical velocity is $0$. $A$ force $F = 10\,N$ acts upward. Since $m = 1\,kg$,the upward acceleration is $a_y = \frac{F - mg}{m} = \frac{10 - 10}{1} = 0\,m/s^2$.Thus,for the next $5\,s$,the body moves horizontally with constant velocity $u_x = 25 \sqrt{3}\,m/s$ and zero vertical acceleration.The horizontal distance covered during this time is $x_2 = u_x 	imes 5 = (25 \sqrt{3}) 	imes 5 = 125 \sqrt{3}\,m$.After $5\,s$,the body falls under gravity from the highest point. The time taken to fall is $t_3 = t_1 = 2.5\,s$.The horizontal distance covered during this time is $x_3 = u_x t_3 = (25 \sqrt{3})(2.5) = 62.5 \sqrt{3}\,m$.The total horizontal range is $R = x_1 + x_2 + x_3 = 62.5 \sqrt{3} + 125 \sqrt{3} + 62.5 \sqrt{3} = 250 \sqrt{3}\,m$.

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