A particle is thrown with a velocity of u m/s. It passes A and B as shown in figure at time `t_(1)` = 1 s and `t_(2)` = 3 s. The value of u is (g = 10 `m//s^(2))`

A particle is moving with velocity v=4t^(3)+3 t^(2)-1 m//s . The displacement of particle in time t=1 s to t=2 s will be

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) )

A projectile is thrown with speed u making angle theta with horizontal at t =0 . It just crosses two points of equal height at time t = 1 s and t = 3 s respectively. Calculate the maximum height attained by it ? (g = 10 m//s^(2) )

A projectile is thrown with speed u making angle theta with horizontal at t =0 . It just crosses two points of equal height at time t = 1 s and t = 3 s respectively. Calculate the maximum height attained by it ? (g = 10 m//s^(2) )

A particle is moving with initial velocity 1m/s and acceleration a=(4t+3)m/s^(2) . Find velocity of particle at t=2sec .

In the arrangement as shown, tension T_(2) is (g=10m//s^(2))

In the arrangement as shown, tension T_(2) is (g=10m//s^(2))

A particle is thrown vertically up from the top of a building of height 20m with initial velocity u. A second particle is released from the same point 1 second later. If both particles reach the ground at the same instant , 3u = m//s. [Take g= 10m//s^(2) ]

A particle is thrown vertically up from the top of a building of height 20m with initial velocity u. A second particle is released from the same point 1 second later. If both particles reach the ground at the same instant , 3u = m//s. [Take g= 10m//s^(2) ]