Question:
A balloon is rising vertically upwards with a velocity of \(10\,\text{m s}^{-1}\). When the balloon is at a height of \(40\,\text{m}\) from the ground, a stone is dropped from it. The time taken by the stone to reach the ground is (Take \(g=10\,\text{m s}^{-2}\)).
A balloon is rising vertically upwards with a velocity of \(10\,\text{m s}^{-1}\). When the balloon is at a height of \(40\,\text{m}\) from the ground, a stone is dropped from it. The time taken by the stone to reach the ground is (Take \(g=10\,\text{m s}^{-2}\)).
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If an object is dropped from a moving balloon, its initial velocity is the same as the velocity of the balloon at that instant.
Updated On: Jul 9, 2026
- \(10\,\text{s}\)
- \(6\,\text{s}\)
- \(8\,\text{s}\)
\(4\,\text{s}\)
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The Correct Option is D
Solution and Explanation
Concept:
When the stone is released, it has the same upward velocity as the balloon.
\[
u=10\,\text{m s}^{-1}
\]
\[
a=-10\,\text{m s}^{-2}
\]
\[
y_0=40\,\text{m}
\]
Step 1: Apply the equation of motion. Taking upward direction as positive, \[ y=y_0+ut+\frac12 at^2 \] At the ground, \[ y=0. \] Hence, \[ 0=40+10t-\frac12(10)t^2 \] \[ 0=40+10t-5t^2 \] \[ t^2-2t-8=0 \]
Step 2: Solve the quadratic equation. \[ (t-4)(t+2)=0 \] \[ t=4,\,-2 \] Rejecting the negative value, \[ t=4\,\text{s} \] Final Answer: \[ \boxed{4\,\text{s}} \]
Step 1: Apply the equation of motion. Taking upward direction as positive, \[ y=y_0+ut+\frac12 at^2 \] At the ground, \[ y=0. \] Hence, \[ 0=40+10t-\frac12(10)t^2 \] \[ 0=40+10t-5t^2 \] \[ t^2-2t-8=0 \]
Step 2: Solve the quadratic equation. \[ (t-4)(t+2)=0 \] \[ t=4,\,-2 \] Rejecting the negative value, \[ t=4\,\text{s} \] Final Answer: \[ \boxed{4\,\text{s}} \]
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