2) Now let us consider the vertical motion.
In vertical direction, the acceleration of the projectile is equal to the free fall acceleration which is constant and always directed downward
a = -gj i.e., ay=-g.
The equation for vertical displacement of the projectile after time t can be written by \(
y = u_y t + \frac{1}
{2}a_y t^2
\),we get
\(
y = (u\sin \theta )t - \frac{1}
{2}gt^2
\) (\(\therefore a_y\)=-g)..........(2)
by substituting the value of t' from (1) as in equation (2)
\(
t = \frac{x}
{{u\cos \theta }}
\)in equation (2)
\(
\begin{gathered}
y = (u\sin \theta )\left( {\frac{x}
{{u\cos \theta }}} \right) - \frac{1}
{2}g\left( {\frac{x}
{{u\cos \theta }}} \right)^2 \hfill \\
\therefore y = \left( {\tan \theta } \right)x - \left( {\frac{g}
{{2u^2 \cos ^2 \theta }}} \right)x^2 \hfill \\
\end{gathered}
\)
The values of g, and u are constants. The above equation is in the form
y=ax-bx2 where a=\(tan\theta\) ;
\(
\therefore \theta = \frac{g}
{{2u^2 \cos ^2 \theta }}
\)
This is the equation of a parabola so the path of the projectile is a parabola