Question:
From a balloon moving upwards with a velocity of 12 ms\(^{-1}\), a packet is released when it is at a height of 65 m from the ground. The time taken by it to reach the ground is (g = 10 ms\(^{-2}\))
From a balloon moving upwards with a velocity of 12 ms\(^{-1}\), a packet is released when it is at a height of 65 m from the ground. The time taken by it to reach the ground is (g = 10 ms\(^{-2}\))
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Always choose sign convention carefully in vertical motion; upward positive simplifies calculations.
Updated On: May 8, 2026
- 5 s
- 8 s
- 4 s
- 7 s
- 10 s
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The Correct Option is A
Solution and Explanation
Concept:
This is a vertical motion problem under gravity. Use equation:
\[
s = ut + \frac{1}{2}at^2
\]
Take upward direction as positive.
Step 1: Write given values. \[ u = +12 \, \text{m/s}, \quad a = -10 \, \text{m/s}^2, \quad s = -65 \, \text{m} \]
Step 2: Substitute into equation. \[ -65 = 12t - 5t^2 \]
Step 3: Rearrange equation. \[ 5t^2 - 12t - 65 = 0 \]
Step 4: Solve quadratic. \[ t = \frac{12 \pm \sqrt{144 + 1300}}{10} = \frac{12 \pm \sqrt{1444}}{10} \] \[ = \frac{12 \pm 38}{10} \]
Step 5: Take positive root. \[ t = \frac{12 + 38}{10} = 5 \, \text{s} \]
Step 1: Write given values. \[ u = +12 \, \text{m/s}, \quad a = -10 \, \text{m/s}^2, \quad s = -65 \, \text{m} \]
Step 2: Substitute into equation. \[ -65 = 12t - 5t^2 \]
Step 3: Rearrange equation. \[ 5t^2 - 12t - 65 = 0 \]
Step 4: Solve quadratic. \[ t = \frac{12 \pm \sqrt{144 + 1300}}{10} = \frac{12 \pm \sqrt{1444}}{10} \] \[ = \frac{12 \pm 38}{10} \]
Step 5: Take positive root. \[ t = \frac{12 + 38}{10} = 5 \, \text{s} \]
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