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  4. A body is projected at an angle θ to the horizontal with kinetic energy Ek . The potential energy at the highest point of the trajectory is

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Q. A body is projected at an angle $\theta $ to the horizontal with kinetic energy $E_{k}$ . The potential energy at the highest point of the trajectory is

NTA AbhyasNTA Abhyas 2020Motion in a Plane

Solution:

Solution
Kinetic energy at highest point $\frac{1}{2} \mathrm{~m}(\mathrm{v} \cos \theta)^2=\mathrm{E}_{\mathrm{k}} \cos ^2 \theta$
Total energy remains constant
$\Rightarrow PE = T . E .- KE _{ k }$
$= E _{ k }- E _{ k } \cos ^{2}$ theta $= E _{ k } \sin ^{2}$ theta