Question

A particle is projected making an angle of $45^{\circ}$ 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)

Kinetic energy of the ball $=K$
and angle of projection $(\theta)=45^{\circ}$.
Velocity of the ball at the highest point
$=v \cos \theta =v \cos 45^{\circ}=\frac{v}{\sqrt{2}}$.
Therefore kinetic energy of the ball
$=\frac{1}{2} m \times\left(\frac{v}{\sqrt{2}}\right)^{2}=\frac{1}{4} m v^{2}=\frac{K}{2}$.