A cricket ball is hit at an angle of 30° to the horizontal with a kinetic energy E. Its kinetic energy when it reaches the highest point is

Question

A cricket ball is hit at an angle of 30° to the horizontal with a kinetic energy E. Its kinetic energy when it reaches the highest point is (A) (E/2) (B) 0 (C) (2E/3) (D) (3E/4)

Tardigrade Admin · Accepted Answer

Given, θ=30° KE =E . We know that, E =(1/2) m v2 ...(i) ⇒ E text total =(1/2) m v2 cos 2 θ =(1/2) m v2 ⋅ cos 2 30° From Eq. (i) , E text total =E cos 2 30° ⇒ E text total =(3 E/4)

Q. A cricket ball is hit at an angle of $3 0^{\circ}$ to the horizontal with a kinetic energy E. Its kinetic energy when it reaches the highest point is

$\frac{E}{2}$

0

$\frac{2 E}{3}$

$\frac{3 E}{4}$

E

Given, $θ = 3 0^{\circ}$
$K E = E .$ We know that,
$E = \frac{1}{2} m v^{2}$ ...(i)
$\Rightarrow E_{total} = \frac{1}{2} m v^{2} cos^{2} θ$
$= \frac{1}{2} m v^{2} \cdot cos^{2} 3 0^{\circ}$
From Eq. (i) , $E_{total} = E cos^{2} 3 0^{\circ}$
$\Rightarrow E_{total} = \frac{3 E}{4}$

Q. A cricket ball is hit at an angle of 30∘ to the horizontal with a kinetic energy E. Its kinetic energy when it reaches the highest point is

2E​

0

32E​

43E​

E

Given, θ=30∘ KE=E. We know that, E=21​mv2 ...(i) ⇒Etotal ​=21​mv2cos2θ =21​mv2⋅cos230∘ From Eq. (i) , Etotal ​=Ecos230∘ ⇒Etotal ​=43E​

Q. A cricket ball is hit at an angle of $3 0^{\circ}$ to the horizontal with a kinetic energy E. Its kinetic energy when it reaches the highest point is

$\frac{E}{2}$

$\frac{2 E}{3}$

$\frac{3 E}{4}$

Given, $θ = 3 0^{\circ}$
$K E = E .$ We know that,
$E = \frac{1}{2} m v^{2}$ ...(i)
$\Rightarrow E_{total} = \frac{1}{2} m v^{2} cos^{2} θ$
$= \frac{1}{2} m v^{2} \cdot cos^{2} 3 0^{\circ}$
From Eq. (i) , $E_{total} = E cos^{2} 3 0^{\circ}$
$\Rightarrow E_{total} = \frac{3 E}{4}$