From a balloon rising vertically upward at $5\,m/s$ a stone is thrown up at $10\,m/s$ relative to the balloon. Its velocity with respect to the ground after $2s$ is. $(g = 10\,m/{s^2})$
(A) $0\,m/s$
(B) $20\,m/s$
(C) $10\,m/s$
(D) $5\,m/s$

Hint: We will find the initial velocity of the balloon which is the addition of the initial velocity of the stone and the balloon as given in the question. Next, we will use the displacement equation for time $t = 2s$ to get the displacement of the stone. We will calculate the velocity using the displacement in the equation ${v^2} = {u^2} + 2aS$.
Formula used: Laws of motion equation
$S = ut + \dfrac{1}{2}a{t^2}$ and
${v^2} = {u^2} + 2aS$.

Complete step by step answer::
It is given that the balloon is rising vertically upwards with a velocity of $5\,m/s$. This is the initial velocity of the balloon and it does not change. From this balloon, we throw a stone at a speed of $10\,m/s$ upwards, in the same direction as that in which the balloon is travelling.
Thus the initial velocity of the stone is $15\,m/s$, since the speed of the balloon will get added to it. Now we have to find the velocity of the balloon at the time $t = 2s$. For this, we will use the laws of motion equation
$S = ut + \dfrac{1}{2}a{t^2}$,
where $S$ is the required displacement.
Substituting the value of $t = 2s$ and the initial velocity of $u = 15\,m/s$ in the above equation, we get $S = (15\,m/s)(2s) + \dfrac{1}{2}( - 10\,m/{s^2}) \times {(2s)^2}$
$ \Rightarrow S = 20\,m$.
Next, we will use
${v^2} = {u^2} + 2aS$, where
$ \Rightarrow {v^2} = {(15\,m/s)^2} + 2( - 10\,m/{s^2})(20\,m) = {(5\,m/s)^2}$
Thus the magnitude of the velocity is
$v = 5\,m/s$
This is the velocity of the stone with respect to the ground after $2s$.

Therefore, the correct answer is option (D).

Note: The initial velocity of the stone needs to be found carefully since the reference frame should be with respect to the ground. Thus we add the velocity of the balloon with respect to the ground with the velocity of the stone with respect to the balloon. Here we will add the two velocities and not subtract, since both the velocities are directed towards the same direction and not opposite or at any other angle.