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

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20 m

30 m

40 m

35 m

answer is C.

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Detailed Solution

Height at which explosion occurs

$H = (u \sin θ) t - \frac{1}{2} g t^{2} = 15 m$

Just before explosion

$V_{x} = u \cos θ = 20 m s^{- 1}$ ,

$V_{y} = (u \sin θ - g t) = 10 m s^{- 1}$

Aplying LCLM,

${\vec{V}}_{2} = 40 \overset{⌢}{i} - 20 \overset{⌢}{j} \Rightarrow h = \frac{V_{2 Y}^{2}}{2 g} = 20 m$

Maximum height reached by the fragment= H+h=35m

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