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 :

Options

A. ` y=x - 5 x^2`
B. ` y = 2 x- 5 x^2`
C. ` 4 y= 2 x - 5 x^2`
D. ` 4 y= 2 x - 25 x^2`

Answer

B

Explanation

Here, initial velocity , ` vec u = hat I + 2 hat j`. Comaring it,
with ` vec u = u_x hat I = u_y hat j`, we have
` u_x = 1 m//s ` and ` u_y = 2 m//s`
Let he projectile at tiem (t) be at location whose position coordinates from origin are (x, y). Then ltbr. ` x= u_x t= 1 . t =t`
and ` y =u_y t- 1/2 gt^2 = 2t - 1/2 xx 10 t^2`
` =2 x - 5 x^2` ( :. t = xx)`.