A body is projected with certain angle from the ground. The motion of the body is described by the equations $x=2t$, $y=3t-4t^2$. Then:

$$\begin{gathered} x=2t \\ t=\frac{x}{2} \\ and y=3\left(\frac{x}{2}\right)-4{\left(\frac{x}{2}\right)}^2 =\frac{3x}{2}-x^2 \\ {u}_x=\frac{dx}{dt}=2 \\ and u_y=\frac{dy}{dt}=3-8t \\ u=\sqrt{{u_x}^2+{u_y}^2}=\sqrt{2^2+{\left(3-8t\right)}^2} \\ =\sqrt{2^2+3^2} \\ =\sqrt{13} \\ Components of acceleration ofthe projectile are given by a_x=\frac{d^{2}x}{d{t}^2}=0 and \\ {a}_y=\frac{d^{2}y}{d{t}^2}=-8 m/s^2.So the acceleration of the projectile is 8 m/s^2. \end{gathered}$$