A gas balloon is going up with a constant velocity of $10$ m/s. When this balloon reached a height of $75$ m, a stone is dropped from it and balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is ________ m. (Take $g=10$ m/s$^2$)

When the stone is dropped, it acquires the initial velocity of the balloon. Taking the upward direction as positive, the initial velocity of the stone is $u = 10$ m/s.

The displacement of the stone when it hits the ground is $S = -75$ m, and its acceleration is $a = -g = -10$ m/s$^2$.

Using the equation of motion $S = ut + \dfrac{1}{2}at^2$:

$-75 = 10t - \dfrac{1}{2}(10)t^2$

$-75 = 10t - 5t^2$

$5t^2 - 10t - 75 = 0$

$t^2 - 2t - 15 = 0$

$(t - 5)(t + 3) = 0$

Since time cannot be negative, $t = 5$ s.

During this time, the balloon continues to move upwards with a constant velocity of $10$ m/s. The distance travelled by the balloon in $5$ s is:

$d = v \times t = 10 \times 5 = 50$ m

The total height of the balloon when the stone hits the ground is:

$H = 75 + 50 = 125$ m

Answer: $125$