A projectile is given an initial velocity of `(hat(i)+2hat(j))` The Cartesian equation of its path is `(g = 10 ms^(-1))` ( Here , `hati` is the unit vector along horizontal and `hatj` is unit vector vertically upwards)

Explanation

`tan theta = (u sin theta)/(u cos theta)=2/1`
The desired equation is
`y = x tan theta - (gx^2)/(2u^(2)cos ^(2) theta)="x"xx2-(10x^2)/(2(sqrt((1)^(2)+(2)^(2)))^(2)(1/sqrt5)^2)`
or `y=2 x - 5x^2`