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(A) The horizontal distance between the two buildings is $d = 100 \; m$. The bullets are fired towards each other with a horizontal velocity $v = 25 \

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after $2 \; s$ at a height of $180 \; m$

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(A) The horizontal distance between the two buildings is $d = 100 \; m$. The bullets are fired towards each other with a horizontal velocity $v = 25 \; m/s$ each.The relative horizontal velocity is $v_{rel} = v_1 + v_2 = 25 + 25 = 50 \; m/s$.The time taken for the bullets to collide is $t = \frac{d}{v_{rel}} = \frac{100}{50} = 2 \; s$.In this time,both bullets fall vertically due to gravity. The vertical displacement $s_y$ is given by $s_y = -\frac{1}{2} gt^2$.Substituting the values: $s_y = -\frac{1}{2} 	imes 10 	imes (2)^2 = -5 	imes 4 = -20 \; m$.The height of the collision from the ground is $H_{collision} = H_{initial} + s_y = 200 - 20 = 180 \; m$.Thus,the bullets collide after $2 \; s$ at a height of $180 \; m$.

Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings $100 \; m$ apart and of the same height of $200 \; m$ with the same velocity of $25 \; m/s$. When and where will the two bullets collide? $(g = 10 \; m/s^2)$

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