JEE MAIN - Physics (2022 - 29th July Morning Shift - No. 5)
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 :
Zero
$$\frac{E}{2}$$
$$\frac{E}{4}$$
E
Објашњење
$$K.E. = E = {1 \over 2}m{v^2}$$
at highest point
$$K.E' = {1 \over 2}m{v^2}{\cos ^2}\theta $$
$$ = {1 \over 2}m{v^2}\left( {{1 \over 4}} \right)$$
$$ = {E \over 4}$$