An aeroplane moving horizontally with a speed of 720 km/h drops a food packet, while flying at a height of 396.9 m. The time taken by a food packet to reach the ground and its horizontal range is (Take g = 9.8 m/sec²)

Correct Option: The correct answer is 9 sec and 1800 m.

To find the time taken by the food packet to reach the ground and its horizontal range, we analyze the motion in horizontal and vertical directions separately.

1. Horizontal Velocity:
The aeroplane is flying horizontally with a speed of 720 km/h. When the food packet is dropped, it inherits this horizontal velocity. First, we convert this velocity into meters per second (m/s):

$u_x=720\times \frac{5}{18}=200\text{ m/s}$

The initial vertical velocity of the food packet is zero ($u_y=0$).

2. Time taken to reach the ground ($t$):
The vertical distance traveled by the packet is $h=396.9\text{ m}$ and the acceleration due to gravity is $g=9.8\text{ m/s}{2}^{}$. Using the second equation of motion for vertical direction:

$h=\frac{1}{2}g{t}^2$

Solving for $t$:

$t=\sqrt{\frac{2h}{g}}$

Substitute the given values:

$t=\sqrt{\frac{2\times 396.9}{9.8}}$

$t=\sqrt{\frac{793.8}{9.8}}=\sqrt{81}=9\text{ seconds}$

3. Horizontal Range ($R$):
The horizontal range is the product of the constant horizontal velocity and the time of flight:

$R=u_x\times t$

$R=200\text{ m/s}\times 9\text{ s}=1800\text{ meters}$

Thus, the time taken is 9 sec and the horizontal range is 1800 m.