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

• A

  $\frac{1}{2} K$
• Correct answer
  B

  $\frac{3}{4} K$
• C

  $\frac{3}{8} K$
• D

  $\frac{5}{8} K$

Solution

Initial kinetic energy is
Final kinetic energy is $\begin{array}{rlrl} K^{\prime} & =\frac{1}{2} m v^2=\frac{1}{2} m\left(u \cos 30^{\circ}\right)^2 & {\left[\because v=u \cos 30^{\circ}\right]} \\ & =\frac{3}{4}\left(\frac{1}{2} m u^2\right)=\frac{3}{4} K & & \text { [From Eq. (i)] }\end{array}$

Hence, the correct answer is option: 'B'.