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Projectile Motion Practice Problems

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  • 0:01 Projectile Motion
  • 0:36 The Equations
  • 4:34 Example 1
  • 6:14 Example 2
  • 9:03 Lesson Summary
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Lesson Transcript
Instructor: Robert Egan
After watching this video lesson, you will know how to use the sets of equations that are used to solve projectile motion problems. Learn how to manipulate them to find the answer you need.

Projectile Motion

Human cannonballs, the path of a football, where an airborne marble will land - all of these are projectile motion problems. Projectile motion refers to the path of an object that has been launched into the air, so the path that a human cannonball takes is a projectile motion problem. Once you solve a projectile motion problem, you will know exactly how long the object will remain airborne and where it will land, along with the path it will take in the air.

The Equations

To solve these projectile motion problems, all you need are two sets of equations. You have two sets of equations because you have both an x component (a horizontal component) and a y component (a vertical component).

The first set, the horizontal set, is this:

Equations

The x represents the horizontal displacement. The v subix stands for the initial horizontal velocity. The v sub fx stands for the final horizontal velocity. The a sub x stand for the horizontal acceleration. And the t stands for time.

So the first equation of this set tells us that our object's horizontal displacement equals its initial horizontal velocity times t plus half the horizontal acceleration times t squared.

The second equation lets us know the final horizontal velocity is equal to the initial horizontal velocity plus the horizontal acceleration times t.

The last equation of this set says that the final horizontal velocity squared equals the initial horizontal velocity squared plus twice the horizontal acceleration times the horizontal displacement.

Even though we have three equations in each set, you may not have to use all three equations. The equations are there to help you, and you can use whichever equations will get the job done. Often, you just need to use one of the equations. For example, most cases of projectile motion have no horizontal acceleration, so we can cancel out any terms with a sub x in it, since a sub x equals zero m/s/s. If the horizontal acceleration is zero, then the second and third equations in this set simply tell you that the initial and final velocities are the same. In this case, only the first equation will help you.

The second set, the vertical set, is basically the same as the first, except it has all the y components. In this set, the y stands for the vertical displacement. The y sub iy stands for the initial vertical velocity. The v sub fy stands for the final vertical velocity. The a sub y stands for the vertical acceleration, and t stands for time.

The first equation tells us that the vertical displacement is equal to the initial vertical velocity times t plus half the vertical acceleration times t squared.

In the next equation, the final vertical velocity equals the initial vertical velocity plus the vertical acceleration times t. The last equation shows that the final vertical velocity squared equals the initial vertical velocity squared plus twice the vertical acceleration times vertical displacement. If your projectile motion problem is happening inside Earth's atmosphere and under Earth's gravity, then your a sub y is equal to -9.8 m/s/s. Note that you have your x and y parts for the velocity. Remember that your velocity is a vector and, unless you are shooting straight out horizontally or vertically, velocity will have both a vertical component and a horizontal component.

To find these components, you will need to apply a bit of trigonometry. To find your initial horizontal velocity, take your initial velocity and multiply it by the cosine of your angle. The angle is the angle the projectile is shot from. To find the initial vertical velocity, you take the initial velocity and multiply it by the sine of the angle. Let's take a look at these equations in action.

Example 1

A paintball is shot horizontally from a height of 0.8 meters with a velocity of 10 m/s. How long does it take before the paintball reaches the floor?

To solve this problem, you first write down all the components that you are given or can figure out. For our horizontal components, we don't know the horizontal displacement, nor do we need it to solve this problem. The initial horizontal velocity is 10 m/s, and the horizontal acceleration is 0, since there is no outside horizontal force acting on the paintball. For our vertical components, the vertical displacement is -0.8 meters because the paintball needs to go down 0.8 meters to reach the ground. The initial vertical velocity is 0, since the paintball is shot horizontally from the gun. And the vertical acceleration, due to gravity, is -9.8 m/s/s.

Now that you've written down everything you know, you need to look through the two sets of equations to see which equations can help you. At this point, your problem becomes less of a physics problem and more of an algebra problem. Looking through your equations, you decide to use the first vertical equation, since you know all the variables in it, except for t, which is what you want to solve for. Actually, this is the only equation you need to find your answer. You plug in your values, then you use algebra to solve for t. This is what you get:

Formula

And you are done! It takes the paintball about 0.4 seconds to reach the ground. That's not even a second!

Example 2

One more problem. Buddy has volunteered to be the human cannonball for the next circus show. He is shot out of a cannon at an angle of 30 degrees with an initial velocity of 80 m/s. What is the maximum height that Buddy reaches? Again, you'll begin by writing down everything you know or can figure out. To figure out your initial horizontal and vertical velocities, you use the cosine and sine functions and multiply them accordingly with the initial velocity.

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