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: Understand the drop.
A plane flies level at $200$ km/hr and at height $980$ m. It releases a bomb. We need the plane's horizontal distance from the target at the moment of release so the bomb lands on it. Take $g = 9.8\ \text{m/s}^2$.

Step 2: Change the speed to SI.
$200\ \text{km/hr} = 200 \times \frac{1000}{3600} = \frac{500}{9}\ \text{m/s}$.

Step 3: Find the time to fall.
The bomb starts with no vertical speed, so $h = \frac{1}{2}g t^2$ gives $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 980}{9.8}} = \sqrt{200} = 10\sqrt{2}\ \text{s}$.

Step 4: Find the horizontal range.
While falling, the bomb keeps its forward speed, so range $= \text{speed} \times \text{time} = \frac{500}{9}\times 10\sqrt{2} = \frac{5000\sqrt{2}}{9}\ \text{m}$.

Step 5: Rewrite in the option form.
Note $\frac{5000\sqrt{2}}{9} = \frac{10^4}{9}\cdot\frac{\sqrt{2}}{2} = \frac{10^4}{9\sqrt{2}}$, since $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

Step 6: State the answer.
The required distance is $\frac{10^4}{9\sqrt{2}}\ \text{m}$.
\[ \boxed{R = \frac{10^4}{9\sqrt{2}}\ \text{m}} \]