Question:
A balloon was moving upwards with a uniform velocity of \(10\ \mathrm{m/s}\). An object of finite mass is dropped from the balloon when it was at a height of \(75\ \mathrm{m}\) from the ground level. The height of the balloon from the ground when object strikes the ground was around :
(takes the value of g as \(10\ \mathrm{m/s^2}\))
A balloon was moving upwards with a uniform velocity of \(10\ \mathrm{m/s}\). An object of finite mass is dropped from the balloon when it was at a height of \(75\ \mathrm{m}\) from the ground level. The height of the balloon from the ground when object strikes the ground was around :
(takes the value of g as \(10\ \mathrm{m/s^2}\))
(takes the value of g as \(10\ \mathrm{m/s^2}\))
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The dropped object starts with the balloon's upward velocity of 10 m/s. Find its time of flight from 75 m to the ground, then add the balloon's rise in that time.
Updated On: Nov 4, 2026
- \(300\ \mathrm{m}\)
- \(200\ \mathrm{m}\)
- \(125\ \mathrm{m}\)
- \(250\ \mathrm{m}\)
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The Correct Option is C
Solution and Explanation
Take upward as positive and the ground as origin. The object leaves the balloon with \(u=+10\ \mathrm{m/s}\) from \(y_0=75\ \mathrm{m}\), with \(a=-10\ \mathrm{m/s^2}\).
When it hits the ground, \(y=0\): \(0 = 75 + 10t - 5t^2\) \(\Rightarrow\) \(t^2 - 2t - 15 = 0\) \(\Rightarrow\) \((t-5)(t+3)=0\) \(\Rightarrow\) \(t = 5\ \mathrm{s}\).
The balloon keeps moving up uniformly at 10 m/s, so in 5 s it rises \(10\times 5 = 50\ \mathrm{m}\).
Height of balloon = \(75 + 50 = 125\ \mathrm{m}\).
Answer: (3) 125 m
When it hits the ground, \(y=0\): \(0 = 75 + 10t - 5t^2\) \(\Rightarrow\) \(t^2 - 2t - 15 = 0\) \(\Rightarrow\) \((t-5)(t+3)=0\) \(\Rightarrow\) \(t = 5\ \mathrm{s}\).
The balloon keeps moving up uniformly at 10 m/s, so in 5 s it rises \(10\times 5 = 50\ \mathrm{m}\).
Height of balloon = \(75 + 50 = 125\ \mathrm{m}\).
Answer: (3) 125 m
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