# 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 $\theta$ in a projectile motion, then at its highest point, the body has no vertical velocity ($v\mathrm{sin}\theta =0$) but it has the horizontal velocity (${v}_{II}=v\mathrm{cos}\theta$).

Thus, the kinetic energy of the body at its highest point will be,

Regards

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