Projectile Motion
To date, all of our lessons have dealt with motion in only one direction. Now, we will look at motion in 2 dimensions, or along both the x and the y axis simultaneously. This is called projectile motion. Solving problems in projectile motion is simply a matter of taking the appropriate quantities and solving for motion in the x axis and the y axis separately. The element that ties the two directions together is time; it is the same for both axes. When solving problems of projectile motion there are several things that you must be aware of:
1) We will ignore air resistance when solving projectile motion problems. If you are curious as to the effect of air resistance, check out the link at the bottom of the page for Projectile Motion with Air Resistance.
2) Motion in the x-axis and y-axis are independent from one another, they
do not affect one another in any way. I
cannot stress this enough - it is the key to solving these problems.
3) It takes the same amount of time for an object to drop vertically as it does for an object to land if projected horizontally. For example, you throw a ball horizontally straight out, that ball that you threw will hit the ground at the same time as if you simply let it fall out of your hand.
4) Have I mentioned that motion in the X axis and motion in the Y axis are independent?
5) The kinematics equations we have learned are the same. We just need to apply them in two different directions. Normally, there will be no acceleration in the x direction. Acceleration in the y direction is due to gravity. Let's try a problem.
A ball is thrown horizontally off of a 100 m building with an initial velocity of 20 m/s. How far away from the building does the ball land?
1. Draw a picture, then establish a table of x and y known values. The value we are looking for is the horizontal distance, which we also call the range. Here is what we know:
Horizontal |
Vertical |
X = |
X_{o} = 100m |
v_{o}=20m/s |
v_{o}=0m/s |
v_{f}=20m/s |
v_{f}= |
a=0 m/s^{2} |
a = -9.8m/s^{2} |
t= |
^{t=} |
Things to note: There is no acceleration in the x direction, and the initial velocity in the y direction is 0 since we are throwing the ball horizontally. The horizontal displacement is what we are looking for. The final velocity in the x direction remains 20 m/s, so range (X) = Vt. We need to find the time the ball is in the air, t, by determining how long it takes the ball to drop 100 m. To do this, I like to use the equation that solves for position as a function of time. This is a quadratic, and we can solve for the variable t very nicely using either the quadratic formula or the programming on our calculators:
we set x_{o} to 100m (initial position), x to 0 m (final position), v_{o} to 0 m/s (initial velocity), a to -9.8 m/s^{2} (acceleration due to gravity) and solve for t. t = 4.5 sec. NOTE: Time will also have a negative root if you solve using the quadratic formula. Ignore it!
Since t is the same in both the horizontal and vertical directions, we can find the range from X = Vt, or range = (20 m/s)(4.5 sec) = 90 m. The ball will land 90 meters from the building.
The NTNU Virtual Physics Laboratory provides several excellent applets that demonstrate principles of Physics. Click on the following to bring up some projectile motion applets:
Interesting Properties of ProjectileMotion
Cannon to Cannon Projectile Motion
Projectile Motion with Air Resistance
Often it is helpful to see what someone else has to say on the topic. Try http://www.physicsclassroom.com/Class/vectors/ and follow the Projectile Motion links for more information.