Question

A bomb is dropped by an aeroplane flying horizontally with a velocity 200 km/hr and at a height of 980 m. At the time of dropping a bomb, the distance of the aeroplane from the target on the ground to hit directly is \( (g = 9.8 \, \text{m/s}^2) \)

• A. \(\frac{\sqrt{2} \times 10^4}{9} \, \text{m}\)
• B. \(\frac{10^4}{9} \, \text{m}\)
• C. \(\frac{10^4}{9\sqrt{2}} \, \text{m}\)
• D. \(\frac{10^4}{18} \, \text{m}\)

Hint:

Convert km/hr to m/s by multiplying by \(\frac{5}{18}\). For height 980 m with \(g = 9.8\), time \(\sqrt{2h/g} = \sqrt{200} = 10\sqrt{2}\) s. Simplify range expression to match given options.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A bomb is dropped from a horizontally flying aeroplane. Initial horizontal velocity \(u = 200\) km/hr, height \(h = 980\) m. We need the horizontal distance from the target (range) so that the bomb hits directly.

Step 2: Key Formula or Approach:
For a projectile dropped from height \(h\) with horizontal velocity \(u\), time of flight \(t = \sqrt{\frac{2h}{g}}\), and horizontal range \(R = u t\).

Step 3: Detailed Explanation:
Convert velocity to m/s: \(200 \, \text{km/hr} = 200 \times \frac{1000}{3600} = \frac{200}{3.6} = \frac{500}{9} \, \text{m/s}\).
Time of flight: \(t = \sqrt{\frac{2 \times 980}{9.8}} = \sqrt{\frac{1960}{9.8}} = \sqrt{200} = 10\sqrt{2} \, \text{s}\).
Range: \(R = \frac{500}{9} \times 10\sqrt{2} = \frac{5000\sqrt{2}}{9} \, \text{m}\).
Rewrite: \(\frac{5000\sqrt{2}}{9} = \frac{5000}{9} \times \sqrt{2} = \frac{10^4}{18} \times \sqrt{2} = \frac{10^4}{9\sqrt{2}}\) (since \(\frac{\sqrt{2}}{18} = \frac{1}{9\sqrt{2}}\)).
Thus \(R = \frac{10^4}{9\sqrt{2}} \, \text{m}\).

Step 4: Final Answer:
Option (C).