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

The correct option is B E/2E'=E cos^2 θ =E cos^2 (45^∘)=E/2 ...

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Standard XII

Physics

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Question

A ball is projected with kinetic energy E at an angle of $45^{\circ}$ to the horizontal. At the highest point during its flight, its kinetic energy will be

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\frac{E}{2}

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\frac{E}{\sqrt{2}}

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E

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Solution

The correct option is B $\frac{E}{2}$
$E^{'} = E c o s^{2} θ = E c o s^{2} (45^{\circ}) = \frac{E}{2}$

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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

The correct option is B E2E′=E cos2 θ=E cos2(45∘)=E2

A ball is projected with kinetic energy E at an angle of $45^{\circ}$ to the horizontal. At the highest point during its flight, its kinetic energy will be

The correct option is B $\frac{E}{2}$
$E^{'} = E c o s^{2} θ = E c o s^{2} (45^{\circ}) = \frac{E}{2}$