`x-y` is the vertical plane as shown in figure. A particle of mass `1kg` is at `(10m, 20m)` at time `t=0`. It is released from rest. Another particles of mass `2kg` is at `(20m, 40m)` at the same instant. It is projected with velocity `(10hati+10hatj)m//s`. After `1s`. Find

v_(x) is the velocity of a particle moving along the x-axis as shown in the figure. If x=2.0m at t=1.0s, what is the position of the particle at t=6.0s ?

A particle of mass 2 kg located at the position (hati + hatj)m has velocity 2(hati - hatj + hatk) m//s . Its angular momentum about Z-axis in kg m^(2)//s is

A particle of mass 2 kg located at the position (hati + hatj)m has velocity 2(hati - hatj + hatk) m//s . Its angualr momentum about Z-axis in kg m^(2)//s is

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 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 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 of mass 1 kg is projected upwards with velocity 60 m/s. Another particle of mass 2 kg is just deopper from a certain height. After 2 kg is just of the particles have colliided with ground, match the following [ Take g= 10 m//s^(2) ]

A particle of mass m is projected up from the bottom of an inclined plane with initial velocity v_(0) at angle 45^(@) with an inclined plane of inclineation 30^(@) as shown in figure. At the same time a small block of same mass m is released from rest at a height h . The particle hits the block at some point on inclined plane. If the particle sticks to the block after collision, the velocity of block parallel to the inclined plane just after collision will be (take g=10m//s^(2) )