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)

$\frac{K}{2}$

Answer 2

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
$=\frac{1}{2}m \times \, \bigg(\frac{v}{\sqrt 2}\bigg)^2 =\frac{1}{4}mv^2=\frac{k}{2}$.