# 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 direction is independent of motion in the direction. So we can write equations of motion for and 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 and direction. It looks like this

Here’s what these equations tell us: acceleration is constant for all time, in both and 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 direction. We’ll call it The minus sign means that it’s directed downward. There is no acceleration in the direction, so

With these, the first two equations become

Next, the velocity equations are

which simplifies to

Then, the position equations are

To summarize, the equations of motion for projectiles are

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

### Examples

Say we throw a ball upward at a 45 angle.

We have these initial conditions: thrown at time thrown from the ground, so and initial velocities are m/s and m/s. Now, our equations of motion are

Questions: (1) How high () does it go? (2) How far () 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 So in order to answer these questions, we need to find some “special” times.

The highest point in the arc is reached when the velocity goes to zero (stops rising and begins falling). This is found by setting like so

which gives

Using m/s for gravity, we get s. This is our first “special” time — the time of maximum height. What is that height? Use the vertical position equation to get

which tells us that the maximum height is 2.5 m.

Next, the end occurs when the ball hits the ground, or when Setting the vertical position equation equal to zero, we get

or

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 and use giving

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

or giving the second root at s. Time is when we threw the ball, so ignore that one. The second “special” time is s, the time of impact.

So to answer how far it went, use the horizontal () position equation, or

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

or m/s. So, it hits with the same velocity it was thrown with, but the sign for is reversed, because at impact, it is moving downward.

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