91Ó°ÊÓ

The initial velocity of the object is $\stackrel{→}{v}$. Since, it is given that the projectile is launched at an angle ${30}^{°},\stackrel{→}{v}$is in a direction that makes an angle ${30}^{°}$with the $x$-axis. Let's resolve $\stackrel{→}{v}$in to $x$and $y$components; ${\stackrel{→}{v}}_x$and${\stackrel{→}{v}}_y$.

The only force acting on the object is the weight. Hence the acceleration of the object $\stackrel{→}{a}$is equal to the gravitational acceleration $g$, which always acts downward (in negative $y$-direction).

Consider the horizontal motion. Since there is no acceleration in the $x$-direction, ${\stackrel{→}{v}}_x$remains unchanged during the projectile.

Now consider the vertical motion. Due to the gravitational acceleration $g$in the negative $y$ direction ${\stackrel{→}{v}}_y$, reduces and becomes zero. Now since there is no vertical velocity the object can't go higher than this point; this is the maximum height of the projectile.

After that the vertical motion of the object is like a free fall.

Let's say the maximum height it reaches is $h$and the final displacement is $s$. The path of the projectile can be sketched as given below.

From the diagram above we can see that, $\stackrel{→}{v}$and $\stackrel{→}{a}$do not become parallel to each other at any point on the trajectory.

From the diagram above we can see that,$\stackrel{→}{v}$ and $\stackrel{→}{a}$become perpendicular to each other at the highest point on the trajectory.