A particle of mass `2m` is projected at an angle of `45^@` with horizontal with a velocity of `20sqrt2m//s`. After `1s` explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. Take `g=10m//s^2`.

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

A particle of mass 1 kg is projected at an angle of 30^(@) with horizontal with velocity v = 40 m/s . The change in linear momentum of the particle after time t = 1 s will be (g = 10 m// s^(2) )

A particle of mass 1 kg is projected at an angle of 30^(@) with horizontal with velocity v = 40 m/s . The change in linear momentum of the particle after time t = 1 s will be (g = 10 m// s^(2) )