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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,$u_x = 100 	imes (4/5) = 80 \ m/s$.The horizontal range $R$ of the projectile is $R = \frac{u^2 \sin 2	heta}{g} = \frac{2 u^2 \sin 	heta \cos 	heta}{g} = \frac{2 	imes 100^2 	imes (3/5) 	imes (4/5)}{10} = 960 \ m$.The horizontal distance to the highest point is $R/2 = 480 \ m$.At the highest point,the momentum is conserved. The initial momentum is $P_i = (2m) u_x = (2m)(80) = 160m$.After the explosion,the first piece $(m)$ comes to rest $(v_1 = 0)$,so $P_f = m(0) + m(v_2) = m v_2$.Equating $P_i = P_f$,we get $160m = m v_2$,so $v_2 = 160 \ m/s$.The second piece travels from the highest point with horizontal velocity $160 \ m/s$ for the remaining time of flight $t = \frac{u \sin 	heta}{g} = \frac{100 	imes (3/5)}{10} = 6 \ s$.The additional horizontal distance covered by the second piece is $d = v_2 	imes t = 160 	imes 6 = 960 \ m$.The total distance from the point of projection is $R/2 + d = 480 + 960 = 1440 \ m$.

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