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 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 is projected at an angle 60^@ with horizontal with a speed v = 20 m//s . Taking g = 10m//s^2 . Find the time after which the speed of the particle remains half of its initial speed.

A particle is projected at an angle 60^(@) with the horizontal with a speed 10m//s . Then, latus rectum is (Take g=10m//s^(2) )

A projectile of mass 3kg is projected with velocity 50m//s at 37^@ from horizontal. After 2s , explosion takes place and the projectile breaks into two parts of masses 1kg and 2kg . The first part comes to rest just after explosion. Find, (a) the velocity of second part just after explosion. (b) maximum height attained by this part. Take g=10m//s^2

A particle of mass 50 g is projected horizontally from the top of a tower of height 50 m with a velocity 20m//s . If g=10m//s^(2) , then find the :-

A particle is thrown vertically upwards. If its velocity is half of the maximum height is 20 m//s , then maximum height attained by it is

A particle is projected at an angle of 30^(@)w.r.t. horizontal with speed 20m//s:( use g=10m//s^(2)) (i) Find the position vector of the particle after 1s . (ii) Find the angle between velocity vector and position vector at t=1s .

A particle of mass m=1kg is projected with speed u=20sqrt(2)m//s at angle theta=45^(@) with horizontal find the torque of the weight of the particle about the point of projection when the particle is at the highest point.