# Mechanics 3: Motion in Two Dimensions — Projectiles

## Motion in Two Dimensions

When something moves in two dimensions (directions) at once, we can treat them independently. What does that mean? Motion in the \(x\) direction is independent of motion in the \(y\) direction. So we can write equations of motion for \(x\) and \(y\) and solve them separately. We finished last chapter by writing the 3 equations of motion — start with acceleration, integrate and add a constant for velocity, integrate again and add a constant for position. Then use the initial conditions to find what are the constants. So, start with the equations of motion from last chapter, and write them for both the \(x\) and \(y\) direction. It looks like this

$$ a_x(t) = a_x \qquad \qquad \qquad \qquad a_y(t) = a_y \\

v_x(t) = a_x t + v_{0x} \qquad \qquad \quad v_y(t) = a_y t + v_{0y} \\

x(t) = \frac12 a_x t^2 + v_{0x} t + x_0 \qquad y(t) = \frac12 a_y t^2 + v_{0y} t + y_0 $$

Here’s what these equations tell us: acceleration is constant for all time, in both \(x\) and \(y\) directions Let’s work through an application to see how these equations are used.

### Projectile Motion

This is a common application with motion in two dimensions (2-D). A projectile is anything thrown, fired, shot, etc. The only acceleration is that of gravity, which is in the \(y\) direction. We’ll call it \(-g.\) The minus sign means that it’s directed downward. There is no acceleration in the \(x\) direction, so \(a_x = 0.\)

With these, the first two equations become

$$ a_x(t) = 0 \qquad \qquad a_y(t) = -g $$

Next, the velocity equations are

$$ v_x(t) = 0t + v_{0x} \qquad \qquad v_y(t) = -gt + v_{0y} $$

which simplifies to

$$ v_x(t) = v_{0x} \qquad \qquad v_y(t) = -gt + v_{0y} \qquad $$

Then, the position equations are

$$ x(t) = v_{0x}t + x_0 \qquad \qquad y(t) = -\frac{1}{2} gt^2 + v_{0y}t + y_0 $$

To summarize, the equations of motion for projectiles are

$$ a_x = 0 \qquad \qquad \qquad a_y = -g \\

v_x = v_{0x} \qquad \qquad v_y(t) = -g t + v_{0y} \\

x(t) = v_{0x} t + x_0 \qquad y(t) = -\frac12 g t^2 + v_{0y} t + y_0 $$

Three of these equations are now constants — they don’t depend on \(t.\) For those, we have removed the notation \((t),\) which means, “is a function of \(t.\)” Let’s work some examples to see how to use all this.

### Examples

Say we throw a ball upward at a 45\(^\circ\) angle. We have these initial conditions: thrown at time \(t=0;\) thrown from the ground, so \(x_0 = 0\) and \(y_0 = 0;\) initial velocities are \(v_{0x} = 7\,\)m/s and \(v_{0y} = 7\,\)m/s. Now, our equations of motion are

$$ a_x = 0 \hskip 1.4in a_y = -g \hskip 0.5in\\

v_x = 7 \hskip 1.2in v_y(t) = -g t + 7 \\

x(t) = 7 t +0 \hskip 0.75in y(t) = -\frac12 g t^2 +7 t + 0 $$

Questions: (1) How high (\(y\)) does it go? (2) How far (\(x\)) does it go? (3) How long till it hits the ground? (4) What is its velocity at impact?

The equations that are not constant depend on \(t.\) So in order to answer these questions, we need to find some “special” times.

The highest point in the arc is reached when the \(y\) velocity goes to zero (stops rising and begins falling). This is found by setting \(v_y(t) = 0,\) like so

$$ 0 = -g t + 7 $$

which gives

$$ gt = 7 $$

Using \(g = 9.8\,\)m/s\(^2\) for gravity, we get \(t = 7 / 9.8 = 0.71\,\)s. This is our first “special” time — the time of maximum height. What is that height? Use the vertical position equation to get

$$y(0.71) = -\frac12 g (0.71)^2 + 7 (0.71)$$

which tells us that the maximum height is 2.5 m.

Next, the end occurs when the ball hits the ground, or when \(y(t) = 0.\) Setting the vertical position equation equal to zero, we get

$$ 0 = -\frac12 g t^2 +7 t$$

or

$$ \frac12 g t^2 – 7 t = 0$$

which is a quadratic equation. Two ways to solve such an equation: factor, or use the quadratic formula. For this case, we just factor out a \(t\) and use \(g=9.8,\) giving

$$ t (4.9t – 7) = 0$$

The roots of this equation are the values of \(t\) that make the equation equal zero. The first term, \(t\) has a root of \(t=0,\) meaning that for time \(t=0,\) the quadratic equation is zero. Then, the second term equals zero when

$$(4.9t – 7) = 0$$

or \(4.9t = 7,\) giving the second root at \(t=1.43\,\)s. Time \(t=0\) is when we threw the ball, so ignore that one. The second “special” time is \(t=1.43\,\)s, the time of impact.

So to answer how far it went, use the horizontal (\(x\)) position equation and we get

$$ x(1.43) = 7 (1.43) $$

or 10 m. How long till it hits the ground is the time-of-flight, or 1.43 s. Finally, velocity at impact is given by both horizontal and vertical equations

$$ v_x = 7 \qquad \qquad v_y(1.43) = -9.8 (1.43) + 7$$

or \(v_y = -7.014\,\)m/s. So, essentially it hits with the same velocity it was thrown with. The sign for \(v_y\) is reversed, because at impact, it’s moving downward.

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