A particle of mass 2m is projected at an angle $45^{0}$ with horizontal with a velocity of $20 \sqrt{2} m / s a$ . After 1 sec, explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. The maximum height attained by the other part from the ground is $(g = 10 m / s^{2})$

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$20 m$

$35 m$

$45 m$

$15 m$

answer is C.

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Detailed Solution

Height attained before explosion

$U_{y} t - \frac{1}{2} {gt}^{2} = (20) (1) - \frac{1}{2} (10) (1) = 20 - 5 = 15 m After 1 \sec V_{x} = 20 m / s v_{y} = v_{y} - gt = 20 - 10 (1) = 10 m / s Apply law of conservation of energy 2 m (V_{x} \bar{i} + V_{y} \bar{j}) = m (O) + m \bar{V} 2 m (20 \bar{i} + 10 \bar{j}) = m \bar{V} \bar{V} = 40 \bar{i} + 20 \bar{j} Height attained after explosion = \frac{V_{y}^{2}}{2 g} = \frac{20 \times 20}{2 \times 10} = 20 m H = 15 + 20 = 35 m$