A particle is projected from the ground level. It just passes through upper end of vertical poles A, B, C of height 24m, 40 m and 36 m respectively. The time taken by the particle to travel from B to C is double of the time taken A to B. Find the maximum height attained by the particle from the ground level. [Take: $g = 10 m / s^{2}$ ]

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answer is 44.17.

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Detailed Solution

$16 = u_{y} t - \frac{1}{2} g t^{2} \times 3$

$12 = u_{y} \times 3 t - \frac{9}{2} g t^{2}$

$36 = 3 g t^{2}$

$t = \sqrt{\frac{12}{g}}$

$16 = u_{y} \sqrt{\frac{12}{g}} - \frac{1}{2} g \frac{12}{g}$

$u_{y} = 22 \sqrt{\frac{g}{12}}$

$H = 24 + \frac{22^{2}}{2 g} \frac{g}{12}$

= $24 + \frac{121}{6}$
= $\frac{144 + 121}{6} = \frac{265}{6}$

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