A balloon is ascending at the rate of $14m{s}^{-1}$ at a height of $98m$ above the ground when the food packet is dropped from the balloon. After how much time and with what velocity does it reach the ground ? Take g = $9.8m{s}^{-2}$

The initial velocity of food packet=velocity of baloon= u=$14\hat{j} m{s}^{-1}$

Acceleration of gravity a =$-9.81\hat{j}m/s^2$

Distance from the ground h =$-98\hat{j}$

let final velocity be v and time taken to reach the ground be t

By second kinematic equation $$\begin{gathered} v^2-u^2=2ah \\ \Rightarrow v=\sqrt{2ah+u^2} \\ \Rightarrow v=\sqrt{2(-9.81)(-98)+{14}^2} \\ \Rightarrow v=-46\hat{j} \end{gathered}$$

Now applying first kinematic equation,

$$\begin{gathered} v=u+at \\ t=\frac{v-u}{a} \\ =\frac{-46-14}{-9.81}=6.11 sec \end{gathered}$$

hence the velocity is $v=-46$ and time $t=6.11 sec$

Option A is correct answer.