The potential energy of a projectile at its highest point is (1/2) th the value of its initial kinetic energy. Therefore its angle of projection is (A) \(30^{\circ}\) (B) \(45^{\circ}\) (C) \(60^{\circ}\) (D) \(75^{\circ}\)

At the highest point of the projectile's trajectory, the vertical component of its velocity equals 0. So the kinetic energy is the square of the horizontal component of velocity divided by 2. On the other hand, the potential energy at the highest point is the gained altitude by the projectile multiplied by the gravitational acceleration (g) and its mass (m).

We know that the speed of the projectile can be written as: \[v = v_0\cos\theta\] The kinetic energy at the highest point will be given by: \[K = \frac{1}{2}mv^2 = \frac{1}{2}m(v_0\cos\theta)^2\] We also know that the height (h) of the highest point can be calculated using the relationship: \[h = \frac{v_0^2\sin^2\theta}{2g}\] The potential energy at the highest point is the product of mass, gravity, and height: \[U = mgh\]

We are given: \[U = \frac{1}{2}K\] Substitute the equations for kinetic and potential energy: \[\frac{1}{2}m(v_0\cos\theta)^2 = mgh\] Cancel out the mass (m) and rearrange the equation: \[\frac{v_0^2\cos^2\theta}{2} = \frac{v_0^2\sin^2\theta}{2g}\] To eliminate one of the trigonometric functions, we can use the identity: \[\sin^2\theta = 1 - \cos^2\theta\] Substitute this identity into the previous equation: \[\frac{v_0^2\cos^2\theta}{2} = \frac{v_0^2(1-\cos^2\theta)}{2g}\] Simplify the equation: \[\cos^2\theta = 1 - \cos^2\theta\] Solve for \(\cos^2\theta\): \[\cos^2\theta = \frac{1}{2}\] Take the square root to get \(\cos\theta\): \[\cos\theta = \frac{1}{\sqrt{2}}\] Finally, determine the angle of projection, \(\theta\): \[\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\] As a result, we obtain the following answer: \[\theta = 45^{\circ}\] So, the correct option is (B) \(45^{\circ}\).