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: \(125\)