A particle of mass 2m is projected at an angle of 45∘ with horizontal with a velocity of 20√2 ms−1. After 1 s explosion takes place and the particle is broken into two equal pieces. As a result of the explosion, one part comes to rest. Find the maximum height (in metre) from initial point of the particle, attained by the other part is [Take g=10 ms−2]

A particle of mass 2m is projected at an angle of 45∘ with horizonta

1.	A particle of mass 2m is projected at an angle of 45∘ with horizontal with a velocity of 20√2 ms−1. After 1 s explosion takes place and the particle is broken into two equal pieces. As a result of the explosion, one part comes to rest. Find the maximum height (in metre) from initial point of the particle, attained by the other part is [Take g=10 ms−2]
Answer» A particle of mass $2 m$ is projected at an angle of $45^{\circ}$ with horizontal with a velocity of $20 \sqrt{2} {ms}^{- 1}$ . After $1 s$ explosion takes place and the particle is broken into two equal pieces. As a result of the explosion, one part comes to rest. Find the maximum height $(in metre)$ from initial point of the particle, attained by the other part is [Take $g = 10 {ms}^{- 2}$ ]

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