If at the highest point of parabolic bath, the kinetic energy is equal to the potential energy, then the angle of projection is equal to:
(A) 30
(B) 45
(C) 60
(D) 70

Hint At the highest point the velocity of the particle in y direction is 0. The x component of the velocity is the same as it was at the beginning of the projectile. So the kinetic energy of the body due to its component is x direction is equal to the potential energy at the highest point.

Complete step by step solution
At the highest point in the trajectory, the velocity pointing in the upward direction becomes 0. But the velocity component in x direction remains the same throughout the motion of the body.
We also know that highest point in a parabolic trajectory is written as:
 \[h\, = \,\dfrac{{{v^2}{{\sin }^2}\theta }}{{2g}}\] ,
The potential energy at this point will be:
 \[mg\dfrac{{{v^2}{{\sin }^2}\theta }}{{2g}}\]
This potential energy should be equal to the kinetic energy of the body, hence
 \[
  mg\dfrac{{{v^2}{{\sin }^2}\theta }}{{2g}}\, = \,\dfrac{1}{2}m{(v\cos \theta )^2} \\
  {v^2}{\sin ^2}\theta \, = \,{v^2}{\cos ^2}\theta \\
  {\tan ^2}\theta = 1 \\
  \theta \, = \,45 \\
 \]

Therefore correct answer is option B

Note The velocity in the x direction will always remain constant as there is no resistive force there, the velocity in y direction reduces to 0 due to gravity and gains its initial velocity moments before touching the ground.