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NEET PHYSICSMedium

Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings (building being 100 m100 \text{ m} apart and being of the same height of 200 m200 \text{ m}) with the same velocity of 25 m/s25 \text{ m/s}. When and where will the two bullets collide? (g=10 m/s2g=10 \text{ m/s}^2)

A

After 2 s2 \text{ s} at a height of 180 m180 \text{ m}

B

After 2 s2 \text{ s} at a height of 20 m20 \text{ m}

C

After 4 s4 \text{ s} at a height of 120 m120 \text{ m}

D

They will not collide.

Step-by-Step Solution

  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 vrel=v1+v2=25+25=50 m/sv_{rel} = v_1 + v_2 = 25 + 25 = 50 \text{ m/s}.
  • Distance to cover d=100 md = 100 \text{ m}.
  • Time t=dvrel=10050=2 st = \frac{d}{v_{rel}} = \frac{100}{50} = 2 \text{ s}.
  1. Vertical Motion: Both bullets have zero initial vertical velocity (uy=0u_y = 0). The vertical distance fallen in time tt is given by h=12gt2h = \frac{1}{2}gt^2 .
  • h=12×10×(2)2=5×4=20 mh = \frac{1}{2} \times 10 \times (2)^2 = 5 \times 4 = 20 \text{ m}.
  1. Height from Ground: The initial height is 200 m200 \text{ m}. Therefore, the height at which they collide is:
  • H=200−20=180 mH = 200 - 20 = 180 \text{ m}.
  1. Conclusion: The bullets collide after 2 s2 \text{ s} at a height of 180 m180 \text{ m}.
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