The initial velocity of a projectile is `(hati+2hatj)m//s` , where `hati and hatj` are unit vectors along the horizontal and vertical directions respectively. Find out the locus of the projectile , taking `g=10 m//s^2`.

Explanation

Horizontal and vertical components of initial velocity are, respectively,
`u_x=1m//s and u_y =2m//s`
Let at t=0 , the initial coordinates of the projectile are (0,0), at t, these are (x,y).
So, `x=u_xt=tm,`
`y=u_yt-1/2"gt"^2=(2t-5t^2)=2x-5x^2`
`therefore y=2x-5x^2` is the locus of the projectile.