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

Dear Student
The expression for kinetic energy is KE= 1/2 mv2
When a body is projected with kinetic energy E and angle θ in a projectile motion, then at its highest point, the body has no vertical velocity (vsinθ=0) but it has the horizontal velocity (vII=vcosθ).

Thus, the kinetic energy of the body at its highest point will be,
 Eh=12mvcosθ2Eh=12mv cos 452 =12 mv2122Eh=E2so, at the highest point the kinetic energy is  E2   .


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