A particle is projected making an angle of $45^{\circ}$ with horizontal having kinetic energy $K$. T

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\

Professor PC Thomas Classes · Answer

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