A helicopter flying horizontally with a speed $100\text{m/s}$ drops a food packet while flying at a height of $20\text{m}$. Calculate the time taken by the food packet to reach the ground and the horizontal range.
$(\text{Take}g=10{\text{m/s}}^2)$

The correct option is B $2\text{s},200\text{m}$
Given:
Horizontal speed of helicopter, $u=100\text{m/s}$
Height of helicopter from ground, $h=20\text{m}$

When the packet is detached from the helicopter, the speed of the packet will be same as horizontal velocity of helicopter.

Therefore, horizontal velocity of food packet is,

$u=100\text{m/s}$

Time of flight for horizontal projectile of food packet is given by

$T=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 20}{10}}=2\text{s}$

Since there is no acceleration in horizontal direction so it will be uniform motion in horizontal direction. Therefore, horizontal range of packet is,

$R=uT=100\times 2=200\text{m}$

Hence, option $\left(\text{C}\right)$ is the correcct answer.