Question 93: 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: