New problem hmmmm

$let's$ $analyse$ $the$ $motion$

Balloon was traveling with an acceleration of $1.25 m/s^{2}$ , therefore the displacement travelled by balloon in 8 seconds is 40m. When stone is dropped then balloon was moving upwards. So,due to inertia stone will move upwards. Therefore, distance of stone must be greater than 40m but,it's displacement will be 40m because it will go to ground and initially it was 40m above the ground.

_from this _

We can conclude that option 1,2 and 3 can't be the answer. So,answer will be ${option 4}$ .

By using the formula $s=ut+\dfrac{at^2}{2}$ , the balloon (which starts at rest) is $1.25\cdot\dfrac{1}{2}\cdot 8^2=40\textnormal{ m}$ above the ground when the stone is released. The stone initially travels upwards (because it has the same initial velocity as the balloon after 6 seconds.

The stone's displacement is $40\textnormal{ m}$ down when it hits the floor, but it has travelled upwards after being released too (so has travelled a distance of more than $40\textnormal{ m}$ ). The answer can therefore only be "reach the ground in $4$ s".

To verify this, we have that $v_\textnormal{stone (initial)}=u_\textnormal{balloon}+a_\textnormal{balloon}\cdot t_\textnormal{flight}=0+1.25\cdot 8=10 \textnormal{ ms}^{-1}$ in the upwards direction.

Then $s=ut+\dfrac{at^2}{2}\implies -40=10t+\dfrac{-9.81\cdot t^2}{2}\iff \dfrac{9.81t^2}{2}-10t-40=0$ .

The quadratic has roots $t=4.05\textnormal{ (3 s.f.)}$ or $t=-2.01\textnormal{ (3 s.f.)}$ .

Since the time taken to hit the floor is positive, the stone reaches the ground in $\color{#20A900}{\boxed{4\textnormal{ s}}}$ .