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

Question

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 (A) Ek (B) Ek cos2 θ  (C) Ek sin2 θ  (D) Ek tan2 θ

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-noaulnqgvqgp.png" /> Kinetic energy at highest point (1/2) mathrm~m( mathrmv cos θ)2= mathrmE mathrmk cos 2 θ Total energy remains constant ⇒ PE = T . E .- KE k = E k - E k cos 2 theta = E k sin 2 theta

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

$E_{k}$

$E_{k} cos^{2} θ$

$E_{k} sin^{2} θ$

$E_{k} tan^{2} θ$

Kinetic energy at highest point $\frac{1}{2} m (v cos θ)^{2} = E_{k} cos^{2} θ$
Total energy remains constant
$\Rightarrow PE = T . E . - K E_{k}$
$= E_{k} - E_{k} cos^{2}$ theta $= E_{k} sin^{2}$ theta

Q. 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

Ek​

Ek​cos2θ

Ek​sin2θ

Ek​tan2θ