A balloon rises vertically upward with constant speed 10 m·s⁻¹. When it reaches a height of 75 m, a stone is released from it. The balloon continues upward at the same speed. Taking g = 10 m·s⁻², what is the height of the balloon above the ground when the stone hits the ground?
Options

Question

A balloon rises vertically upward with constant speed 10 m·s⁻¹. When it reaches a height of 75 m, a stone is released from it. The balloon continues upward at the same speed. Taking g = 10 m·s⁻², what is the height of the balloon above the ground when the stone hits the ground?

Options

G1 Physics · Accepted Answer

At the instant of release, the stone has the same upward velocity as the balloon: u = +10 m·s⁻¹. Take upward as positive. For the stone, initial height = 75 m, acceleration = -10 m·s⁻². When it reaches the ground, y = 0. Thus, 0 = 75+10t-(1 ÷ 2)(10)t². So, 0 = 75+10t-5t². Divide by 5: 0 = 15+2t-t². Therefore, t²-2t-15 = 0. Factor: (t-5)(t+3) = 0. The physical solution is t = 5 s. During these 5 s, the balloon rises by Δh = 10×5 = 50 m. Hence its height is 75+50 = 125 m. Correct option: D.