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

Question

A ball is projected with kinetic energy E, at an angle of 60° to the horizontal. The kinetic energy of this ball at the highest point of its flight will become : (A) Zero (B) (E/2) (C) (E/4) (D) E

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c8554b20d1461214fb8d70a1cf5f09c2-.png" /> E =(1/2) mu 2 At Highest point, Velocity V = u cos 60°=( u /2) ∴ K . E at topmost point =(1/2) m (( u /2))2=( E /4)

Q. A ball is projected with kinetic energy E, at an angle of $6 0^{\circ}$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :

Zero

$\frac{E}{2}$

$\frac{E}{4}$

$E$

$E = \frac{1}{2} m u^{2}$
At Highest point, Velocity $V = u cos 6 0^{\circ} = \frac{u}{2}$
$∴ K . E$ at topmost point $= \frac{1}{2} m (\frac{u}{2})^{2} = \frac{E}{4}$

Q. A ball is projected with kinetic energy E, at an angle of 60∘ to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :

Zero

2E​

4E​

E

E=21​mu2 At Highest point, Velocity V=ucos60∘=2u​ ∴K.E at topmost point =21​m(2u​)2=4E​

Q. A ball is projected with kinetic energy E, at an angle of $6 0^{\circ}$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :

$\frac{E}{2}$

$\frac{E}{4}$

$E$

$E = \frac{1}{2} m u^{2}$
At Highest point, Velocity $V = u cos 6 0^{\circ} = \frac{u}{2}$
$∴ K . E$ at topmost point $= \frac{1}{2} m (\frac{u}{2})^{2} = \frac{E}{4}$