A ball is hit at 45° to the horizontal with a kinetic energy Ek . The kinetic energy at the highest point is

Question

A ball is hit at 45° to the horizontal with a kinetic energy Ek . The kinetic energy at the highest point is (A)  Ek  (B)  (Ek/2)  (C)  (Ek/√2)  (D) zero

Tardigrade Admin · Accepted Answer

Key Idea: At the highest point vertical component of velocity is zero. Kinetic energy is possessed due to velocity. If m is mass of ball, then kinetic energy is

E_{k} = \frac{1}{2} m u^{2}

At the highest point or path only horizontal component of velocity exists, hence kinetic energy is,

E_{k} = \frac{1}{2} m (u cos 45^{\circ})^{2} = \frac{1}{2} m (\frac{u}{2})^{2}

∴

E_{k} = \frac{1}{2} (\frac{1}{2} m u^{2}) = \frac{E _{k}}{2}

Q. A ball is hit at $45^{\circ}$ to the horizontal with a kinetic energy $E_{k}$ . The kinetic energy at the highest point is

$E_{k}$

$\frac{E _{k}}{2}$

$\frac{E _{k}}{2}$

zero

Q. A ball is hit at 45∘ to the horizontal with a kinetic energy Ek​ . The kinetic energy at the highest point is

Ek​

2Ek​​

2​Ek​​

zero

Q. A ball is hit at $45^{\circ}$ to the horizontal with a kinetic energy $E_{k}$ . The kinetic energy at the highest point is

$E_{k}$

$\frac{E _{k}}{2}$

$\frac{E _{k}}{2}$