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A particle is suspended from a fixed point by a string of length $5 m$ . It is projected from the equilibrium position with such a velocity that the string slackens after the particle has reached a height $8 m$ above the lowest point. Find at what height the particle can rise further _____m.

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answer is 0.96.

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Detailed Solution

As shown in figure at point $B, T = 0$

Thus we have $\frac{m v^{2}}{l} = m g \cos θ$

Or $v = \sqrt{g l \cos θ} = \sqrt{29.4}$

Or $= 5.42 m / s$

$h = \frac{v^{2} \sin^{2} θ}{2 g} = \frac{29 A \times {(4 / 5)}^{2}}{2 \times 9.8} = 0.96 m$

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