1.

A body is projected at an angle 45° to the horizontal with K.E. ‘E’. The potential energy at the highest point of flight is :

Answer»

Zero
`E/4`
`E/2`
`(3E)/4`

Solution :Potential ENERGY at HIGHEST point =`mgH_(MAX)`
`E_(p)=mgxx(u^(2)sin^(2)theta)/(2g)=1/2mu^(2)sin^(2)45^@=E/2`


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