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(B) Given: Initial speed $u = 20 \, m/s$,angle $	heta = 30^{\circ}$,height $h = 40 \, m$.Vertical component of initial velocity: $u_y = u \sin 30^{\c

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$40 \sqrt{3}$

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(B) Given: Initial speed $u = 20 \, m/s$,angle $	heta = 30^{\circ}$,height $h = 40 \, m$.Vertical component of initial velocity: $u_y = u \sin 30^{\circ} = 20 	imes \frac{1}{2} = 10 \, m/s$.Horizontal component of initial velocity: $u_x = u \cos 30^{\circ} = 20 	imes \frac{\sqrt{3}}{2} = 10\sqrt{3} \, m/s$.Using the equation of motion in the vertical direction: $S_y = u_y T + \frac{1}{2} a_y T^2$.Taking the downward direction as negative,$S_y = -40 \, m$ and $a_y = -g = -10 \, m/s^2$.$-40 = 10T - \frac{1}{2} 	imes 10 	imes T^2$.$-40 = 10T - 5T^2$.Dividing by $-5$: $T^2 - 2T - 8 = 0$.$(T - 4)(T + 2) = 0$.Since time cannot be negative,$T = 4 \, s$.Horizontal range $R = u_x 	imes T = 10\sqrt{3} 	imes 4 = 40\sqrt{3} \, m$.

The figure shows a projectile thrown with a speed $u = 20 \, m/s$ at an angle of $30^{\circ}$ with the horizontal from the top of a building $40 \, m$ high. The horizontal range of the projectile is ........... $m$.

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