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PhysicsW.E.P.Work kinetic energy theoremEasy2 minPYQ_2020

PhysicsEasysingle choice

A ball whose kinetic energy is $E$ , is thrown at an angle of $45 °$ with the horizontal. Its kinetic energy at the highest point of its flight will be:

Options:

Answer:

Solution:

Kinetic energy is E $\frac{1}{2} m v^{2} = E \Rightarrow v = \sqrt{\frac{2 E}{m}}$ Velocity at highest point $= v \cos 45 °$ $∴ v \cos 45 ° = \sqrt{\frac{2 E}{m}} . \frac{1}{\sqrt{2}} = \sqrt{\frac{E}{m}}$
[ $∵$ at the highest point there is only horizontal velocity] $∴$ K.E. at highest point $= \frac{1}{2} \times m \frac{E}{m} = \frac{E}{2}$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Work kinetic energy theorem

⏱ 2mℹ️ Source: PYQ_2020

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