Solve this:
Q.5. 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
(a) y = x - $5x^2$
(b) y = 2x - $5x^2$
(c) 4y = 2x - $5x^2$
(d) 4y = 2x - 25$x^2$

Dear student,
Initial Velocity (u) = i^{^}+2j^{^} m/s

Hence horizontal component of initial velocity (u_x) = i^ m/s

Therefore |u_x| = 1m/s

Also angle of projection (0) = tan^-1(u_y/u_x)= tan^-1(2/1)

Therefore tan0 = 2

Trajectory or path of a projectile is given by: y = xtan0 - gx² / 2u_x²

Puting ux = 1 and tan0 = 2, we get,

y = 2x - gx² / 2

or y = 2x - 5x²

Regards