A particle travels according to the equation `y=x-(x^(2)/(2))`. Find the maximum height it acheives (x and y are in metre)

Explanation

Compare it with the equation of trajectory of a projectile. `y=x tan theta-(gx^(2))/(2u^(2)cos^(2)theta)`
`tan theta =1 rArr theta=45^(@) & (g)/(2u^(2)cos ^(2)theta)=(1)/(2)rArr u^(2)=(g)/(1//2)=2g. H=(u^(2)sin^(2)theta)/(2g)=(2gxx(1)/(2))/(2g)=(1)/(2)m`
or
At maximum height `(dy)/(dx)=1-(2x)/(2)=0rArrx=1` so `y_("max")=1-(1)/(2)=(1)/(2)m`