An object of mass 2m is projected with a spee | vedclass

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(C) The initial horizontal velocity of the object is $u_x = u \cos 	heta$. Given $\sin 	heta = 3/5$,then $\cos 	heta = 4/5$. So,$u_x = 100 	imes (

Vedclass · Accepted Answer

$1440$

Vedclass · Answer

(C) The initial horizontal velocity of the object is $u_x = u \cos 	heta$. Given $\sin 	heta = 3/5$,then $\cos 	heta = 4/5$. So,$u_x = 100 	imes (4/5) = 80 \ m/s$.At the highest point,the vertical velocity is $0$,and the horizontal velocity is $80 \ m/s$. The momentum of the system at the highest point is $P = (2m) 	imes 80 = 160m$.After the explosion,the first piece $(m)$ comes to rest,so its velocity is $0$. Let the velocity of the second piece $(m)$ be $v$. By conservation of linear momentum: $160m = m(0) + m(v)$,which gives $v = 160 \ m/s$.The time taken to reach the highest point is $t = (u \sin 	heta) / g = (100 	imes 3/5) / 10 = 6 \ s$. The horizontal distance covered to reach the highest point is $x_1 = u_x 	imes t = 80 	imes 6 = 480 \ m$.The second piece travels from the highest point to the ground in time $t = 6 \ s$ with a constant horizontal velocity of $160 \ m/s$. The additional horizontal distance covered is $x_2 = v 	imes t = 160 	imes 6 = 960 \ m$.The total distance from the point of projection is $x_1 + x_2 = 480 + 960 = 1440 \ m$.

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