2D Motion

Projectile Motion

One of the convenient aspects of displacement, velocity, and acceleration all being vector quantities, is that each of the three components for the x-, y-, and z-directions are all independent of one another. In other words, an acceleration in the z-direction will only affect the velocity in the y-direction. The x and y-components of the velocity remain unchanged. This permits us to look at a projectile in flight as three separate kinematic problems, rather than trying to tackle the whole problem at once.

Projectiles in 2D

If we consider a projectile in flight and ignore the complication of wind resistance, we can see that it will always be a 2D problem. The x-axis we orient along the projectile's horizontal path, and the y-axis in the vertical direction. For this case, the acceleration in the horizontal direction is zero, $a_x=0$, and in the vertical direction is the acceleration due to gravity, $a_y=-g$. Applying these conditions to the kinematic equations we derived in previously in Kinematics, we can create five equations that describe the motion of any projectile.

Horizontal Motion:

(1)
\begin{equation} x_f = x_i + v t \end{equation}

Vertical Motion:

(2)
\begin{align} y_f = y_i + v_i t - {\scriptstyle \frac{1}{2}} g t^2 \end{align}
(3)
\begin{align} y_f = y_i + \left( \frac {v_i + v_f} {2} \right) t \end{align}
(4)
\begin{equation} v_f = v_i - g t \end{equation}
(5)
\begin{align} v_f^2 = v_i^2 - 2 g \left( y_f - y_i \right) \end{align}

When using these five equations, it will be quite common to solve one direction for time in order to gain information about the motion in the other direction. Consider this example:

Example:
A stunt car runs off of a 15-m high building at a speed of 22 m/s. How far away from the building does the car land?

Since the initial height of the car is know, the time it takes to fall can be calculated. Keep in mind that its horizontal speed is irrelevant to the time it takes to fall, only the vertical velocity, which is zero here, matters.

$v_{y,i} = 0$
$y_i = 15 m$
$y_f = 0$
Looking at Eq. 2, $0 = y_i - {\scriptstyle \frac{1}{2}} g t^2$
Solving for t, $t = \sqrt { \frac {2 * y_i} {g} } = 1.75 s$
Inserting time into Eq. 1, $x_f = x_i + v_x t = 22 \frac {m} {s} 1.75 s = 38.5 m$

Very often, the initial velocity is given as a vector in magnitude and direction form rather than as separate horizontal and vertical components as in the above problem. In this case, one must first resolve the vector into the horizontal and vertical components before beginning the problem. If the direction for the velocity is given as an angle above the horizontal, then the initial x- and y- components are given by

(6)
\begin{align} v_x = v_i \cos \left( \theta \right) \end{align}
(7)
\begin{align} v_{yi} = v_i \sin \left( \theta \right) \end{align}

Exercise:
A shotput leaves the thrower's hand at a height of 2.20 m and with a velocity of 10.6 m/s at 40 degrees above the horizontal. How far away from the thrower does the shotput land?

If the projectile is fired over flat, level ground, then we can derive a simple equation that will yield the total range of the projectile, so long as we ignore air resistance. If we start with $x_i = 0$, then Eq. 1 reduces to $x_f = v_x t$. If the initial velocity is known, then all that's needed to find the total distance, $x_f$, isthe total time, $t$. To find the total time, we focus on the fact that if the projectile takes off and lands at the same height, then $y_f = y_i$ so that Eq. 2 becomes,

(8)
\begin{align} 0 = v_{yi} t - {\scriptstyle \frac{1}{2}} g t^2 \end{align}

We can divide both sides by $t$ since we know $t \neq 0$ and then solve.

(9)
\begin{align} t = \frac {2 v_{yi}} {g} \end{align}

Inserting this into our equation for the total range and making the substitutions of $v_x$ and $v_{yi}$ given in Eqs. 6 and 7 gives us our final solution.

(10)
\begin{align} x_f = v_i \cos \left( \theta \right) \cdot \frac {2 v_i \sin \left( \theta \right)} {g} x_f = \frac {{v_i}^2} {g} 2 \cos \left( \theta \right) \sin \left( \theta \right) \end{align}

Recalling that $2 \cos \left( \theta \right) \sin \left( \theta \right) = \sin \left( 2 \theta \right)$

(11)
\begin{align} x_f = \frac {{v_i}^2} {g} \sin \left( 2 \theta \right) \end{align}

If we examine Eq. 11 closely, we can see that for a given initial velocity, there is one angle that yields a maximum range, 45 degrees. Lets look at an example:

Projectiles in 3D

Relative Velocity

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