The final velocity fo a particles falling freelly under graavity is given by `V^(2) - U^(2) = 2GX` WHERE `x` is the distance covered. IF `V = 18` kmph. `g = 1000 cm s^(-2), x = 120 cm` then `u = ...ms^(-1)`

The initial velocity of a particle is given by u^(2) = v^(2) - 2gx where x is the distance covered. IF u = 18 kmh^(-1), g = 1000 cm//s^(2) x = 150 cm then v = ......... m//s

What would be the distance covered by the particle falling freely in 1s? (g = 10 ms^(-2) )

The velocity of a particle is given by v=u_(0) + g t+ 1/2 at^(2) . If its position is x =0 at t=0 , then what is its displacement after t=1 s ?

The initial velocity of a particle- is u and acceleration (f) is uniform. Final velocity and distance covered in the interval t are v and s respectively. Show that the velocity of the particle at half - distance is more than the velocity for half-time.

The distance travelled by a particle in n^(th) second is S_n =u+a/2 (2n-1) where u is the velocity and a is the acceleration. The equation is

The distance travelled by a particle in n^(th) second is S_n =u+a/2 (2n-1) where u is the velocity and a is the acceleration. The equation is

The acceleration of a particle in ms^(-1) is given by a=3t^(2)+2t+2 where timer is in second If the particle starts with a velocity v=2ms^(-1) at t=0 then find the velocity at the end of 2s.

(a) The motion of the particle in simple harmonic motion is given by x = a sin omega t . If its speed is u , when the displacement is x_(1) and speed is v , when the displacement is x_(2) , show that the amplitude of the motion is A = [(v^(2)x_(1)^(2) - u^(2)x_(2)^(2))/(v^(2) - u^(2))]^(1//2) (b) A particle is moving with simple harmonic motion is a straight line. When the distance of the particle from the equilibrium position has the values x_(1) and x_(2) the corresponding values of velocity are u_(1) and u_(2) , show that the period is T = 2pi[(x_(2)^(2) - x_(1)^(2))/(u_(1)^(2) - u_(2)^(2))]^(1//2)