Question

A particle is projected at ${60}^{00}$ to the horizontal with a kinetic energy K. The kinetic energy at the highest point is-

• A. K
• B. zero
• C. K /4
• D. K /2

Answer 1

The correct option is C K /4

$\begin{array}{c}Letubetheinitialvelocity\\ \therefore initialkineticenergyis\\ K=\frac{1}{2}m{u}^2\\ Atthemaximumheightvelcityisucos{60}^{\circ }=\frac{u}{2}\\ \therefore KineticenergyatthemaximumheightisKE=\frac{1}{2}m{\left(\frac{u}{2}\right)}^2\\ =\frac{K}{4}\\ Hence,thecorrectanswerisC.\end{array}$