Question 94: Work, Energy and Power & Circular Motion

Question

A ball is projected with kinetic energy E, at an angle of $60^{\circ}$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :

Accepted Answer

$K.E. = E = {1 \over 2}m{v^2}$
at highest point
$K.E' = {1 \over 2}m{v^2}{\cos ^2}	heta $
$ = {1 \over 2}m{v^2}\left( {{1 \over 4}} ight)$
$ = {E \over 4}$

Answer

Zero

Answer

$\frac{E}{2}$

Answer

$\frac{E}{4}$

Answer

E

Work, Energy and Power & Circular Motion

A ball is projected with kinetic energy E, at an angle of 60^{\circ} to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :

Choose the correct answer: