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$$^2$$)

Solution :

Given :

Initial height of stone,

$$u_y = 75\text{ m}$$

Initial velocity of stone upward,

$$u = 10\text{ m s}^{-1}$$

Acceleration due to gravity,

$$a = -10\text{ m s}^{-2}$$

When stone is released, it has same upward velocity as balloon.

Taking upward direction positive,

Equation of motion :

$$y = ut + \frac{1}{2}at^2$$

When stone hits ground :

$$y = -75\text{ m}$$

Therefore,

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

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

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

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

$$t = 5\text{ s}$$

During this time, balloon continues moving upward with constant speed :

$$v = 10\text{ m s}^{-1}$$

Distance moved upward by balloon :

$$s = vt$$

$$= 10 \times 5$$

$$= 50\text{ m}$$

Therefore, height of balloon when stone hits ground :

$$= 75 + 50$$

$$= 125\text{ m}$$

Final Answer :

$$125\text{ m}$$