A packet is released from a balloon which is moving upwards when the balloon is at a height $100m$ above ground. The packet reaches the ground in $8s$. Speed of the balloon when the packet is released is $(g = 10\dfrac{m}{{{s^2}}})$
(A)$28\dfrac{m}{s}$
(B)$20\dfrac{m}{s}$
(C)$15\dfrac{m}{s}$
(D)$10\dfrac{m}{s}$

Hint If a body is moving with a velocity $v$ and if an object is dropped from it without applying any force on it the object also attains the velocity $v$. But in this process, the object being dropped does not acquire the acceleration of the body from which it is dropped.
Formula used
$h = ut + \dfrac{1}{2}a{t^2}$
Where $h$ is the distance which a body is covering in time $t$ with an initial velocity $v$ and acceleration is $a$.

Complete Step-by-step solution
The data given in the question is,
$t = 8s$
$h = 100m$
$g = 10\dfrac{m}{{{s^2}}}$
We know that,
If a body is moving with a velocity $v$ and if an object is dropped from it without applying any force on it the object also attains the velocity $v$. But in this process, the object being dropped does not acquire the acceleration of the body from which it is dropped.
Hence the initial velocity of the packet will be the velocity with which the balloon moves upward.
And
$\Rightarrow$ $h = ut + \dfrac{1}{2}a{t^2}$
Where $h$ is the distance which a body is covering in time $t$ with an initial velocity $v$ and acceleration is $a$.
Using the data given and the formula known,
$\Rightarrow$ $h = ut + \dfrac{1}{2}g{t^2}$
$100 = 8u + \dfrac{1}{2}(10)({8^2})$
$\Rightarrow$ $8u = 100 - \dfrac{{10 \times 64}}{2}$
On solving further,
$\Rightarrow$ $8u = - 220$
$u = - 27.5\dfrac{m}{s}$
$u \approx - 28\dfrac{m}{s}$
Initially, our balloon was moving in an upward direction due to which the packet also started to move in an upward direction and here we took our sign convention that downward direction is positive due to which the initial velocity of packet came out to be negative which indicated that the packet was moving in an upward direction.

Therefore the correct answer is (A)$28\dfrac{m}{s}$ .

Note
We can solve this question by taking an upward direction as positive. In that case, our height would get a negative sign as the body is travelling that distance in the downward direction, acceleration would also become negative as the acceleration due to gravity is pointing in the downward direction and our initial velocity will become positive.