A particle is projected from the ground such that its horizontal range is three times its maximum height. The ratio of its kinetic energy at the highest point of its trajectory to its initial kinetic energy is

Step-by-step solution

Let the initial velocity be u and the angle of projection be . Given that the horizontal range R is three times the maximum height H : R = 3H Using the standard expressions for range and maximum height: u^2 2 g = 3 ( u^2 ^2 2g ) 2 u^2 g = 3 u^2 ^2 2g = 4 3 From this, we can find : = 3 5 The initial kinetic energy is: K_ initial = 1 2 m u^2 At the highest point, the vertical component of velocity is zero, and the horizontal component is u . The kinetic energy at the highest point is: K_ top = 1 2 m (u )^2 = K_ initi