A bomb is dropped on an enemy post by an aeroplane flying with a horizontal velocity of 60 km/hr and at a height of 490 m. How far the aeroplane must be from the enemy post at time of dropping the bomb, so that it may directly hit the target. (g = 9.8 m/s²)

The correct option is 500/3 m.

Step-by-step Explanation:

Let's analyze the motion of the bomb after it is dropped from the aeroplane. The bomb undergoes projectile motion. It has an initial horizontal velocity equal to that of the aeroplane, and its initial vertical velocity is zero.

1. Convert the horizontal velocity to standard units (m/s):
The horizontal velocity of the aeroplane is:
$u=60\text{km/hr}$
To convert km/hr to m/s, we multiply by $\frac{5}{18}$:
$u=60\times \frac{5}{18}=\frac{300}{18}=\frac{50}{3}\text{m/s}$

2. Find the time taken by the bomb to reach the ground:
The vertical motion of the bomb is a free fall under gravity from a height $h=490\text{m}$.
Using the second equation of motion for the vertical direction:
$h=u_{y}t+\frac{1}{2}g{t}^2$
Since the initial vertical velocity $u_y=0$:
$h=\frac{1}{2}g{t}^2$
Substitute the given values ($h=490\text{m}$ and $g=9.8{\text{m/s}}^2$):
$490=\frac{1}{2}\times 9.8\times t^2$
$490=4.9t^2$
$t^2=\frac{490}{4.9}=100$
$t=10\text{s}$

3. Calculate the horizontal distance traveled by the bomb:
During this time of 10 seconds, the bomb travels horizontally with a constant velocity $u$.
The horizontal distance (range) $R$ is given by:
$R=u\times t$
Substitute the values of $u$ and $t$:
$R=\frac{50}{3}\times 10=\frac{500}{3}\text{m}$

Therefore, the aeroplane must drop the bomb when its horizontal distance from the enemy post is exactly 500/3 m so that it hits the target directly.