A balloon is ascending at the rate of$14m/s$ 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 will the food packet reach the ground? Take g = $9.8m{s}^{-2}$

The initial velocity of the food packet=velocity of baloon= u=$-14 m/s$
The negative sign indicates that the food packet was traveling against gravity initially.

Acceleration of gravity a =$9.8 m/s^2$

Distance from the ground s =$98m$

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

By the second kinematic equation $$\begin{gathered} v^2-u^2=2as \\ \Rightarrow v=\sqrt{2as+u^2} \\ \Rightarrow v=\sqrt{2(9.8)\left(98\right)+{(-14)}^2} \\ \Rightarrow v=46 \end{gathered}$$

Now applying the first kinematic equation,

$$\begin{gathered} v=u+at \\ t=\frac{v-u}{a} \\ =\frac{46-(-14)}{9.8}=6.11 sec \end{gathered}$$

hence the velocity is $46m/s$and time $t=6.11 sec$

Option A is the correct answer.