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 )

Step-by-step solution

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 + 1 2 at^2 : -75 = 10t - 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 upwar