Question

A particle is projected making angle 45⁰ with horizontal having kinetic energy K. The kinetic energy at highest point will be

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

Answer 1

Answer: (B)

$\frac{K}{2}$

Answer 2

Kinetic energy at the highest point $K' = K\cos^{2}\theta = K\cos^{2}45^{o} = K/2$