Projectile Motion
8.1.1. State the independence of the vertical and horizontal components of motion for a projectile in a uniform field.

When a projectile is shot in a uniform gravitational the horizontal and vertical components of motion are independent, that is the horizontal motion does not affect the vertical motion. No matter how fast the projectile moves vertically it can not affect the horizontal motion. Projectile motion can be seen as the superposition of horizontal and vertical motion. Projectile motion can be described using the uniformly accelerated motion equations.

Special case for horizontal motion:

Since there are no horizontal forces (excluding air resistance) acting on a projectile, the horizontal velocity is constant, i.e. there is no horizontal acceleration. Thus the equations of UAM become:

\begin{equation} v = u + {a} t \end{equation}
\begin{equation} v = u \end{equation}

So the initial horizontal velocity is equal to the final horizontal velocity. This greatly simplifies many of the equations:

\begin{align} s={u+v \over 2}t \end{align}
\begin{align} s={u+u \over 2}t \end{align}
\begin{equation} s=ut \end{equation}

The third equation: Simplifies to the same equation.

\begin{align} s=ut + \frac{1}{2}at^2 \end{align}

Simplifies to the same equation:

\begin{equation} s = ut \end{equation}

The fourth equation:

\begin{equation} v^2 = u^2 + 2as \end{equation}

Simplifies to:

\begin{equation} v^2 = u^2 \end{equation}


\begin{equation} v =u \end{equation}

From these equations we can see that horizontal motion for projectiles is very simple and is described by only two equations:

\begin{equation} s=ut \end{equation}
\begin{equation} v=u \end{equation}

Vertical and horizontal motion is not completely separate, they are connected in time. There is not a separate time for horizontal motion or vertical motion. This allows us to solve or partially solve motion in direction and then apply some of that information to the motion in the other direction.

8.1.2 Describe the trajectory of projectile motion as parabolic in the absence of friction.

In the absence of friction the displacement of a projectile can be described with the following two equations:

\begin{equation} s=u_h t \end{equation}
\begin{align} s_v = u_v t + \frac{1}{2}at^2 \end{align}

Where the subscripts h and v refer to horizontal and vertical respectively. Solving the first equation for time and substituting into the second equation we get:

\begin{align} s_v = u_v \left (s_h \over u_h \right ) + \frac{1}{2}a \left (s_h^2 \over u_h^2 \right ) \end{align}

Rearranging we get:

\begin{align} s_v = \left (a \over 2u_v^2 \right) s_h^2 + \left (u_v \over u_h \right) s_h \end{align}

If we where to then plot the vertical displacement vs. the horizontal displacement the plot would be parabola.

\begin{equation} y = ax^2 +bx \end{equation}

This can be clearly seen if we set

\begin{equation} y= s_v \end{equation}
\begin{align} a = \left (a \over 2u_v^2 \right) \end{align}
\begin{align} b = \left(u_v \over u_h \right) \end{align}
\begin{equation} x = s_h \end{equation}
\begin{equation} x^2 = s_h^2 \end{equation}
A check for understanding. Which ball will hit the ground first? And why?

Want to add to or make a comment on these notes? Do it below.

Add a New Comment
or Sign in as Wikidot user
(will not be published)
- +
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License