A balloonist releases a bag from a balloon, rising at $40m/s$ at a time when the balloon is $100m$ above the ground. If $g=10m/s^2$, then the bag reaches the ground in

The correct option is C

$10s$

Step 1: Given Data

Since the balloon is going upwards against the gravity, the initial velocity is $u=-40m/s$

Height from which packet is released $h=100m$

Final velocity $v=0$

Acceleration due to gravity $g=10m/s^2$

Step 2: Formula Used

According to Newton's second law of motion

$S=ut+\frac{1}{2}a{t}^2$

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

$\Rightarrow 2h=2ut+g{t}^2$

$\Rightarrow g{t}^2+2ut-2h=0$

Step 3: Calculate the time taken

Upon substituting the values we get,

$10t^2+2\times \left(-40\right)t-2\left(100\right)=0$

$\Rightarrow t^2-8t-20=0$

$\Rightarrow t^2-10t+2t-20=0$

$\Rightarrow t\left(t-10\right)+2\left(t-10\right)=0$

$\Rightarrow \left(t-10\right)\left(t+2\right)=0$

$\Rightarrow t=10,-2$

Since time cannot be negative,

$t=10s$

Hence, the correct answer is option (C).