A particle is projected horizontally from the top of a tower with a velocity `v_(0)`. If v be its velocity at any instant, then the radius of curvature of the path of the particle at that instant is directely proportional to

A ball is projected horizontal from the top of a tower with a velocity v_(0) . It will be moving at an angle of 60^(@) with the horizontal after time.

A ball is projected horizontal from the top of a tower with a velocity v_(0) . It will be moving at an angle of 60^(@) with the horizontal after time.

A particle is projected with a velocity 8 m//sec at an angle 45^(0) , with the horizontal.What is the radius of curvature of the trajectory of the particle at the instant of '1/4th' of the time of ascent.

A particle is projected with a velocity bar(v) = ahat(i) + bhat(j) . Find the radius of curvature of the trajectory of the particle at (i) point of projection (ii) highest point .

A particle moves with initial velocity v_(0) and retardation alphav , where v is velocity at any instant t. Then the particle

A particle moves with initial velocity v_(0) and retardation alphav , where v is velocity at any instant t. Then the particle

A particle is projected horizontally from a tower with velocity V_(0) . Wind is blowing in opposite direction and is providing a constant horizontal acceleration a_(0) . If particle strikes the ground vertically then height of building is

A particle is projected from the ground at an angle of 60^(@) with horizontal at speed u = 20 m//s. The radius of curvature of the path of the particle, when its velocity. makes an angle of 30^(@) with horizontal is : (g=10 m//s^(2)