1.

A ball whose kinetic energy is E is projected at an angle of 45^(@) to the horizontal. The kinetic energy of the ball at the highest point of its flight willbe:

Answer»

E
`Esqrt(2)`
`E//2`
zero

Solution :At the HIGHEST point of trajectory velocity` =UPSILON cos theta`
K.E. at the throw `=(1)/(2)m upsilon^(2)cos^(2) 45^(@)`
K.E. at the highest point = `(1)/(2)mupsilon^(2)cos^(2)45^(@)`
`EXX((1)/(sqrt2))^(2)=(E)/(2)`
Hence choice is (c)


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