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

Option 1)

E

Option 2)

E/\sqrt{2}

Option 3)

E/2

Option 4)

Zero

Answers (2)

As we learnt in

Kinetic energy -

k= \frac{1}{2}mv^{2}

- wherein

m\rightarrow mass

v\rightarrow velocity

kinetic Energy is never negative

 

 

Initial kinetic energy E=\frac{1}{2}mv^{2}

Kinetic energy at highest point =\frac{1}{2}m\left (v cos45^{o} \right )^{2}

 

    =\frac{1}{2}\left (\frac{1}{2}mv^{2} \right )=\frac{E}{2}

Correct answer is 3.    


Option 1)

E

This is an incorrect option.

Option 2)

E/\sqrt{2}

This is an incorrect option.

Option 3)

E/2

This is the correct option.

Option 4)

Zero

This is an incorrect option.

N neha

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