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If y = x - (1)/(2)x^(2) is the equation...

If `y = x - (1)/(2)x^(2)` is the equation of a trajectory , find the time of flight.

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We have `y=x-1/2x^(2)-x(1-x//2)`
If y=0, then either x=0 or x=2
Hence the arange of the motion is 2.
For half the range x=1 then `y=1//2`
Hence maximum height attained `=1//2`
Time to reach maximum height `t=sqrt((2y)/g)=sqrt(1/g)`
Time of flight `T=2t=2sqrt(1/g)` i.e. `T=2/(sqrt(g))`
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