A projectile is given an initial velocity of ` ( hat(i) + 2 hat (j) ) m//s`, where ` hat(i)` is along the ground and `hat (j)` is along the vertical . If ` g = 10 m//s^(2) `, the equation of its trajectory is :

Explanation

Here, ` u cos theta = 1 and ` u sin theta = 2`
So, ` tan theta = (u sin theta)/( u cos theta) = 2/1`
and ` cos theta 1/( sqrt5) , u= sqrt(1^1 + 2^2 ) = sqrt 5`
. ltbtgt The equation of path of projectile is
` y=x tan theta- ( gx^2)/(2 u^2 cos^2 theta)`
` = x (2) - ( 10x^2)/(2 xx ( sqrt 5)^2 xx (1// sqrt 5)^2 ) = 2 x - 5 x^2`.