# Projectile motion  (Page 2/16)

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Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so ${A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ and ${A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ are used. The magnitude of the components of displacement $\mathbf{s}$ along these axes are $x$ and $\mathrm{y.}$ The magnitudes of the components of the velocity $\mathbf{v}$ are ${v}_{x}=v\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ and ${v}_{y}=v\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\theta ,}$ where $v$ is the magnitude of the velocity and $\theta$ is its direction, as shown in [link] . Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

$\text{Horizontal Motion}\left({a}_{x}=0\right)$
$x={x}_{0}+{v}_{x}t$
${v}_{x}={v}_{0x}={v}_{x}=\text{velocity is a constant.}$
$\text{Vertical Motion}\left(\text{assuming positive is up}\phantom{\rule{0.25em}{0ex}}{a}_{y}=-g=-9.\text{80}{\text{m/s}}^{2}\right)$
$y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t$
${v}_{y}={v}_{0y}-\text{gt}$
$y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\mathrm{gt}}^{2}$
${v}_{y}^{2}={v}_{0y}^{2}-2g\left(y-{y}_{0}\right)\text{.}$

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time $t$ . The problem solving procedures here are the same as for one-dimensional kinematics    and are illustrated in the solved examples below.

Step 4. Recombine the two motions to find the total displacement $\mathbf{\text{s}}$ and velocity $\mathbf{\text{v}}$ . Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing $A=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}}$ and $\theta ={\text{tan}}^{-1}\left({A}_{y}/{A}_{x}\right)$ in the following form, where $\theta$ is the direction of the displacement $\mathbf{s}$ and ${\theta }_{v}$ is the direction of the velocity $\mathbf{v}$ :

Total displacement and velocity

$s=\sqrt{{x}^{2}+{y}^{2}}$
$\theta ={\text{tan}}^{-1}\left(y/x\right)$
$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}$
${\theta }_{v}={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right)\text{.}$

## A fireworks projectile explodes high and away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of $75.0º$ above the horizontal, as illustrated in [link] . The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

Strategy

Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which ${a}_{x}=0$ and ${a}_{y}=–g$ . We can then define ${x}_{0}$ and ${y}_{0}$ to be zero and solve for the desired quantities.

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