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 ($g = 9.8 m/s^2$)

• A. $\frac{\sqrt{2 \times 10^4}}{9} m$
• B. $\frac{10^4}{9} m$
• C. $\frac{10^4}{9\sqrt{2}} m$
• D. $\frac{10^4}{18} m$

Answer 1

Answer: (C)

$\frac{10^4}{9\sqrt{2}} m$

Answer 2

Here's how to solve this problem:

1. Convert the speed:
   Convert 200 km/hr to m/s:
   $200 \text{ km/hr} = 200 \times \frac{5}{18} = \frac{500}{9} \text{ m/s}$

2. Determine the time of fall:
   Use the formula for vertical motion under constant acceleration:
   $h = \frac{1}{2}gt^2$, where $h$ is the height, $g$ is the acceleration due to gravity, and $t$ is the time.

   $980 = \frac{1}{2} \times 9.8 \times t^2$
   $t^2 = \frac{980 \times 2}{9.8} = 200$
   $t = \sqrt{200} = 10\sqrt{2} \text{ s}$

3. Calculate the horizontal distance:
   Since the horizontal velocity is constant, the horizontal distance $d$ is given by:
   $d = u \times t = \frac{500}{9} \times 10\sqrt{2} = \frac{5000\sqrt{2}}{9} \text{ m}$

4. Match the result to the options:
   Notice that $\frac{10^4}{9\sqrt{2}} = \frac{10000}{9\sqrt{2}} = \frac{5000 \times 2}{9\sqrt{2}} = \frac{5000\sqrt{2}}{9} \text{ m}$

Therefore, the correct answer is $\frac{10^4}{9\sqrt{2}} \text{ m}$