Question

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

• A. $\frac{K}{2}$
• B. $K$
• C. $zero$
• D. $\frac{K}{4}$

Answer 1

Answer: (D)

$\frac{K}{4}$

Initial K.E, $K=\frac{1}{2}m{u}^2$

At highest point $u_v=0$ where $u_v$ is velocity in vertical direction

$\therefore v=u_h=u\cos {60}^{\circ }=\frac{u}{2}$ where $v$ is velocity at height point

$\therefore K.E=\frac{1}{2}m{\left(\frac{u}{2}\right)}^2=\frac{1}{8}m{u}^2=\frac{K}{4}$