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(B) Let the initial horizontal velocity be $u$. The vertical motion is unaffected by the wind. Time of flight $T$ is given by $H = \frac{1}{2} g T^2$,

Vedclass · Accepted Answer

$\frac{80}{3} \ m/s^2$

Vedclass · Answer

(B) Let the initial horizontal velocity be $u$. The vertical motion is unaffected by the wind. Time of flight $T$ is given by $H = \frac{1}{2} g T^2$,where $H = 40 \ m$ and $g = 10 \ m/s^2$. $40 = \frac{1}{2} 	imes 10 	imes T^2 \implies T^2 = 8 \implies T = 2\sqrt{2} \ s$. The final vertical velocity is $v_y = gT = 10 	imes 2\sqrt{2} = 20\sqrt{2} \ m/s$. The final horizontal velocity is $v_x = u - aT$,where $a$ is the acceleration due to wind. Since the particle hits the base of the tower,the horizontal displacement $S_x = uT - \frac{1}{2} a T^2 = 0$. Thus,$u = \frac{1}{2} a T$. Substituting $u$ into the expression for $v_x$: $v_x = \frac{1}{2} a T - a T = -\frac{1}{2} a T$. The angle with the horizontal is $37^\circ$,so $	an 37^\circ = \frac{|v_y|}{|v_x|}$. $\frac{3}{4} = \frac{20\sqrt{2}}{\frac{1}{2} a (2\sqrt{2})} = \frac{20\sqrt{2}}{a\sqrt{2}} = \frac{20}{a}$. $a = \frac{20 	imes 4}{3} = \frac{80}{3} \ m/s^2$.

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