Question

An object is thrown at a 60° angle of horizontal with K kinetic force. Its kinetic energy at the highest point of track?​

Answer 1

Answer:

At the highest point of motion ,the vertical component of velocity becomes zero. Only horizontal component of velocity exists and it remains constant through out the motion(if air resistance is neglected).

As kinetic energy is given, it's velocity can be easily derived.

E = (1/2)×M×v^2.

=> V= √(2E/M).

The horizontal component of velocity will be(given angle of projection 60°) vcos60°.

=> V (horizontal) = √(2E/M)×(1/2) = √(E/M×1/2).

=> K.E at heighest point = 1/2×M× v ^2 .( Here v is horizontal velocity).

=> K.E = E/4.

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Answer 2

$\huge\star{\tt{\underline{\underline{\red{Answer}}}}}\star$

E=½mv²

v=$\sqrt{2\frav{E}{M}}$

Horizontal Component =vcos60°

V=$\sqrt{2\frav{E}{M}}$x½=$\sqrt{\frav{E}{M2}}$

THEREFORE

KE=¼