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(C) The net force on the projectile is given by $\vec{F}_{net} = m \frac{d\vec{v}}{dt}$.In the horizontal direction,the only force acting is the visco

Vedclass · Accepted Answer

$170$

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(C) The net force on the projectile is given by $\vec{F}_{net} = m \frac{d\vec{v}}{dt}$.In the horizontal direction,the only force acting is the viscous drag force $F_x = -c v_x$.Thus,$m \frac{dv_x}{dt} = -c v_x$.Rearranging the terms,we get $\frac{dv_x}{v_x} = -\frac{c}{m} dt$.Integrating both sides from $t = 0$ to $t$ and $v_x = v_{0x}$ to $v_x$,we get $\ln \left( \frac{v_x}{v_{0x}} ight) = -\frac{c}{m} t$.So,$v_x = v_{0x} e^{-(c/m)t}$.Given $m = 200 \ g = 0.2 \ kg$ and $c = 0.1 \ kg/s$,the ratio $c/m = 0.1 / 0.2 = 0.5 \ s^{-1}$.Thus,$v_x = v_{0x} e^{-0.5t}$.Since $v_x = \frac{dx}{dt}$,we have $x = \int_0^t v_{0x} e^{-0.5t} dt = v_{0x} \left[ \frac{e^{-0.5t}}{-0.5} ight]_0^t = 2 v_{0x} (1 - e^{-0.5t})$.Given $v_0 = 270 \ m/s$ and angle $	heta = 60^{\circ}$,$v_{0x} = v_0 \cos 60^{\circ} = 270 	imes 0.5 = 135 \ m/s$.At $t = 2 \ s$,$x = 2 	imes 135 	imes (1 - e^{-0.5 	imes 2}) = 270 	imes (1 - e^{-1}) = 270 	imes (1 - 1/2.7) = 270 	imes (1.7 / 2.7) = 100 	imes 1.7 = 170 \ m$.

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