Question

a particle is projected making angle 45° with horizontal having kineric energy k. the kinetic energy at hightest point will be....?

Answer 1

Given:
let v be initial velocity.
K.E= E =1/2 mV²
VELOCITY OF PARTICLE AT HIGHEST POINT = vcos 45
kinetic energy of particle at highest point is =1/2 m(vcos45)²
=1/2(mv²)(1/2)
=1/2 E



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Answer 2

K.E= E =1/2 mV²

VELOCITY OF PARTICLE AT HIGHEST POINT = vcos 45

kinetic energy of particle at highest point is =1/2 m(vcos45)²

=1/2(mv²)(1/2)

=1/2 E