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: Convert the speed.
$200\ \text{km/hr} = 200\times\tfrac{1000}{3600} = \tfrac{500}{9}\ \text{m/s}$.

Step 2: Find the fall time.
The bomb drops from height $980\ \text{m}$, so $t = \sqrt{\tfrac{2h}{g}} = \sqrt{\tfrac{2(980)}{9.8}} = \sqrt{200} = 10\sqrt2\ \text{s}$.

Step 3: Find the horizontal distance.
\[ R = u t = \frac{500}{9}\times10\sqrt2 = \frac{5000\sqrt2}{9}. \]

Step 4: Rewrite the form.
Since $\tfrac{5000\sqrt2}{9} = \tfrac{10^4}{9\sqrt2}$, \[ \boxed{R = \frac{10^4}{9\sqrt2}\ \text{m}} \]