Question

2. A particle is projected at 60° to the horizontal with a kinetic energy K. The kinetic energy at the
highest point is:
(AIEEE-2007)
1) K
2) Zero
3) K/4
4) K/2​

Answer 1

Answer:

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Explanation:

Answer 2

Ans is K/4

Explanation:

In Projectile motion at highest point projectile has no vertical component of velocity..so the only is horizontal component = u cos theta

initial kinetic energy

$k = \frac{1}{2} mv {}^{2}$

At highest point v = v cos theta

= v cos (60°)

= v/2

Kinetic energy at highest point

KE = 1/2 m (v/2)^2

=1/2 mv^2/4

=K/4

Hope it helps..

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