From the top of a tower of height 40 m,a bal | vedclass

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(D) The vertical component of the initial velocity of the ball is:$u_y = u \sin 30^{\circ} = 20 	imes \frac{1}{2} = 10 \ m/s$Taking the upward direct

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$4$

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(D) The vertical component of the initial velocity of the ball is:$u_y = u \sin 30^{\circ} = 20 	imes \frac{1}{2} = 10 \ m/s$Taking the upward direction as negative and the downward direction as positive,the displacement $h = 40 \ m$ is in the downward direction.Using the equation of motion $h = u_y t + \frac{1}{2} g t^2$:$40 = -10 t + \frac{1}{2} 	imes 10 	imes t^2$$40 = -10 t + 5 t^2$$5 t^2 - 10 t - 40 = 0$Dividing by $5$:$t^2 - 2 t - 8 = 0$Factoring the quadratic equation:$t^2 - 4 t + 2 t - 8 = 0$$t(t - 4) + 2(t - 4) = 0$$(t + 2)(t - 4) = 0$Since time cannot be negative,$t = 4 \ s$.

List-$I$	List-$II$
$(i) \ a \bar{a}$	$(A) -\frac{\pi}{3}$
$(ii) \ \arg \left(\frac{1}{\bar{a}}\right)$	$(B) -i \sqrt{3}$
$(iii) \ a-\bar{a}$	$(C) \frac{2i}{\sqrt{3}}$
$(iv) \ \operatorname{Im}\left(\frac{4}{3a}\right)$	$(D) 1$
	$(E) \frac{\pi}{3}$
	$(F) \frac{2}{\sqrt{3}}$

From the top of a tower of height $40 \ m$,a ball is projected upwards with a speed of $20 \ m/s$ at an angle $30^{\circ}$ to the horizontal. The ball will hit the ground in time ......... $\sec$ (Take $g = 10 \ m/s^2$)

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