A particle is thrown at angle θ to cover maximum range with kinetic energy ‘K’. What is its energy at the highest point?
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A particle is thrown at angle θ to cover maximum range with kinetic energy ‘K’. What is its energy at the highest point?
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Given, Kinetic energy K
, Let velocity of particle isv.
⇒
v=
m
2K
⇒
2
1
mv
2
= K
Velocity in horizontal direction
=
vcos(θ)=
m
2K
cos(θ)Velocity in vertical direction
=
vsin(θ)=
m
2K
sin(θ)
At highest point, Vertical velocity is zero only constant horizontal velocity is present.
Kinetic energy at highest point
=
2
1
m(vcos(θ))
2
=
2
1
×
m×
m
2K
cos
2
(θ)=
Kcos
2
(θ)
Answer: K*cos^2(x)
Explanation: bat highest point usinx=0
therefore K.E. = 1/2 m U^2[COS^2(x)]
therefore K.E. = K*cos^2(x)