Question

A particle is thrown at angle θ to cover maximum range with kinetic energy ‘K’. What is its energy at the highest point?

Answer 1

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 2

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)