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WBJEE · Maths · Application of Derivatives

From a balloon rising vertically with uniform velocity \(\mathrm{ft} / \mathrm{sec}\) a piece of stone is let go. The height of the balloon above the ground when the stone reaches the ground after \(4 \mathrm{sec}\) is \(\left[\mathrm{g}=32 \mathrm{ft} / \mathrm{sec}^2\right]\)

  1. A \(220 \mathrm{ft}\)
  2. B \(240 \mathrm{ft}\)
  3. C \(256 \mathrm{ft}\)
  4. D \(260 \mathrm{ft}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(256 \mathrm{ft}\)

Step-by-step Solution

Detailed explanation

When stone is let go, its velocity \(=-v\) (downwards) Let, it was at a height h. \(\therefore h=-v t+\frac{1}{2} gt^2\) \(\begin{aligned} &=-v \times 4+\frac{1}{2} \times 32 \times 4^2 \\ &=-4 v+256 \end{aligned}\) Balloon reaches \(4 \mathrm{vtt}\) in 4 seconds \(\therefore\)…