When a particle is thrown horizontally from certain height above the ground with an initial velocity u, the speed of the particle at any time t before striking the ground is given by
(A) gt
(B) `1/2 g``t^(2)`
(C) `sqrt(u^(2)+g^(2)t^(2))`
(D) `sqrt(u^(2)-g^(2)t^(2))`

Distance travelled by particle at any time t is given by S=t^(2)+6t+8 Speed of particle at any time t is

A body is projected from the ground with a initial velocity vecu an angle theta with the horizontal. Speed with which it strikes the ground (A) u costheta (B) u sintheta (C) u (D) none

A particle thrown up vertically from the ground at reaches its highest point at t=t_0 =The displacement of the particle between t=( 7t_0)/(8) and t=t_0 is S. Then (u^(2))/(Sg) =

A particle is thrown vertically upward from the ground with some velocity and it strikes the ground again in time 2 s. The maximum height achieved by the particle is : (g=10 m//s^(2))

A particle is projected at an angle theta from ground with speed u (g = 10 m/ s^2 )

A particle is projected horizontally from a tower with velocity u towards east. Wind provides constant horizontal acceleration (g)/(2) towards west. If particle strikes the ground vertically the then :

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A particle with initial velocity u is moving with uniform acceleration (a). The distance moved by the particle in the t^(th) second is given by S=u+(1)/(2)a(2t-1) The dimensions of S must be

A particle is projected horizontally with a speed v_(0) at t = 0 from a certain point above the ground . What is the magnitude of tangential acceleration of the particle at t = t_(1) is ( Assume that the particle does not strike the ground between the time interval t = 0 to t = t_(1) )

The displacement of a body is given by 2 s= g t^(2) where g is a constant. The velocity of the body at any time t is :