A ball having kinetic energy KE, is projected at an angle of 6 0 ∘ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight ?

Initial K.E, K.E. = 2 1 mu 2 Speed at heighest point V = u cos 6 0 ∘ = 2 u ∴ KE 2 = 2 1 m ( 2 u ) 2 = 4 1 × 2 1 mu 2 = 4 KE

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PhysicsW.E.P.MiscellaneousEasy2 minQB

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A ball having kinetic energy KE, is projected at an angle of $6 0^{\circ}$ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight ?

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Answer:

Solution:

Initial K.E, $K.E. = \frac{1}{2} mu^{2}$ Speed at heighest point $V = u cos 6 0^{\circ} = \frac{u}{2} ∴ KE_{2} = \frac{1}{2} m (\frac{u}{2})^{2} = \frac{1}{4} \times \frac{1}{2} mu^{2} = \frac{KE}{4}$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Miscellaneous

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