Question

a particle is projected to 45° with horizontal having kinetuc energy K. what is kinetic energy at heightest point

Answer 1

Answer:

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

at maximum height the vertical(usin@) component of the velocity becomes zero

horizontal component =ucos@

initial velocity =u

initial kinetic energy =1/2mu²=k

k at maximum height=1/2(mucos@)²

@=45°

k`=1/2(mucos45)²

=1/2mu²(1/2)

=k/2

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