A ball is projected with kinetic energy E at an angle of 45° to the horizontal. At the highest point during its flight, its kinetic energy will be

Question

A ball is projected with kinetic energy E at an angle of 45° to the horizontal. At the highest point during its flight, its kinetic energy will be (A) Zero (B) (E/2) (C) (E/√2) (D) E

Tardigrade Admin · Accepted Answer

E'=E cos 2 θ=E cos 2(45°)=(E/2)

Q. A ball is projected with kinetic energy $E$ at an angle of $4 5^{\circ}$ to the horizontal. At the highest point during its flight, its kinetic energy will be

Zero

$\frac{E}{2}$

$\frac{E}{2}$

E

$E^{^{'}} = E cos^{2} θ = E cos^{2} (4 5^{\circ}) = \frac{E}{2}$

Q. A ball is projected with kinetic energy E at an angle of 45∘ to the horizontal. At the highest point during its flight, its kinetic energy will be

Zero

2E​

2​E​

E

E′=Ecos2θ=Ecos2(45∘)=2E​

Q. A ball is projected with kinetic energy $E$ at an angle of $4 5^{\circ}$ to the horizontal. At the highest point during its flight, its kinetic energy will be

$\frac{E}{2}$

$\frac{E}{2}$

$E^{^{'}} = E cos^{2} θ = E cos^{2} (4 5^{\circ}) = \frac{E}{2}$