Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings (b

Question

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

Accepted Answer

1. **Horizontal Motion:** Since both bullets are fired horizontally towards each other, we can use relative velocity to find the time of collision. 
   - Relative velocity $v_{rel} = v_1 + v_2 = 25 + 25 = 50 	ext{ m/s}$.
   - Distance to cover $d = 100 	ext{ m}$.
   - Time $t = \frac{d}{v_{rel}} = \frac{100}{50} = 2 	ext{ s}$.
2. **Vertical Motion:** Both bullets have zero initial vertical velocity ($u_y = 0$). The vertical distance fallen in time $t$ is given by $h = \frac{1}{2}gt^2$ .
   - $h = \frac{1}{2} 	imes 10 	imes (2)^2 = 5 	imes 4 = 20 	ext{ m}$.
3. **Height from Ground:** The initial height is $200 	ext{ m}$. Therefore, the height at which they collide is:
   - $H = 200 - 20 = 180 	ext{ m}$.
4. **Conclusion:** The bullets collide after $2 	ext{ s}$ at a height of $180 	ext{ m}$.

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