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. K
• B. Zero
• C. $K/4$
• D. $K/2$

Answer 1

Answer: (C)

$K/4$

let u be the initial velocity.

initial kinetic energy is$K=\left(\frac{1}{2}\right)m{u}^2$

at the maximum height velocity is$ucos60=\left(\frac{u}{2}\right)$

kinetic energy at the maximum height is$KE=\left(\frac{1}{2}\right)m\left(\frac{u}{2}\right)²$

$=\left(\frac{1}{4}\right)k$

Answer C: