1.

a body is projected such that k E at thetop most position is half of intinal kE what is the angle of projection with the horizontal

Answer» LET Velocity at the botton be uu

Then Kinetic Energy at the BOTTOM is KE=1/2mu2KE=1/2mu2

Now at the top only x-component of velocity is present

Let velocity at top be v

Then v=root{(UCOS theta)^2}[Theta is the angle]

According to problem,

(1/2)(1/2mu^2)=1/2mv^2

=>(1/2)u^2=u^2cos^2(theta)

=>cos theta=1/root 2

theta=45 degrees


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