A particle is projected at an angle 30^ with horizontal having kinetic energy K . The kinetic energy of the particle at highest point is.

Step-by-step solution

Let the initial velocity of the particle be u . The initial kinetic energy is given by K = 1 2 mu^2 . At the highest point of the trajectory, the vertical component of the velocity becomes zero, while the horizontal component remains constant. The horizontal component of the velocity is v_x = u , where = 30^ . The kinetic energy at the highest point K' is given by K' = 1 2 m(v_x)^2 = 1 2 m(u 30^ )^2 . Substituting 30^ = 3 2 , we get K' = 1 2 mu^2 ( 3 2 )^2 . Since K = 1 2 mu^2 , we have K' = K 3 4 = 3 4 K . Answer: