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

Explanation

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