2-Dimensional Kinematics = 2 1-Dimensional Questions

2-Dimensional Kinematics = 2 1-Dimensional Questions

We have all seen a similar parabolic arc to the one here, and have been asked similar questions in our 2-dimensional kinematics units: How far will the ball travel? How high will it get? How fast will it be travelling when it hits the ground? Many of these questions are fairly straight forward in the 1-dimensional case, where the boy is simply throwing the ball straight up in the air, but students get tripped up by the 2-dimensional question. What’s important to understand is that, as long as we can break a vector up into components, a 2-dimensional question is really just two overlapping 1-dimensional questions!

Getting Rid Of That Angle

Often times, I try to encourage students to get rid of the thing that is confusing about a questions, and the beauty of math and algebra, is that often times we can. In this case, the part of the question that trips people up is the initial velocity and angle, and what they should do with that information. In fact, dealing with the initial velocity is a fairly simple 3-step process:

1) Break the initial velocity up into its component vectors


3) Ask yourself, is the question a vertical question (gravity included) or a horizontal question (no gravity)?

Take the initial velocity and angle depicted below for instance. What we want to do is break that up into component vectors, as step 1 above suggests. To do that, we simply use basic trig equations:


y = 25 m/s * sin(300)

y = 12.5 m/s

x = 25 m/s * cos(300)

x = 21.65 m/s


Now that we have the vector broken up into components, step 2 suggests that we ignore the initial velocity and angle. This is in fact the best way to approach these questions. Once we’ve broken the original vector into its components, now any question either becomes a “vertical” or “horizontal” question, and you simply need to use the right component to answer the right question.

How High Will The Ball Go?

To illustrate how we now use these components, let’s solve one of the most commonly asked 2-D kinematics question for our example: How high will the ball go? First we need to ask ourselves, “is this a vertical or a horizontal question?” Well, height is a vertical distance, so it would seem that this is a vertical question. Since it is a vertical question, we will want to use the equations from our kinematics unit that include gravity (ex. vf2 = vi2 – 2ad). By using this equation and solving for “d”, we get:

d = (vf2 – vi2)/(2a)

The last thing to note, is you may remember in the 1-dimensional case, when solving for maximum height, we would use the above equation and set vf = 0. This is because, at the top of the objects flight, before it begins to fall, the ball is in fact motionless with no velocity. This element is one of the most confusing when we move to the 2-D case, as the ball never “stops” moving. However, by breaking the question up in to separate vertical and horizontal questions, the logic is the same. In the vertical direction, the ball in fact does stop at the highest point (same as 1-D), and thus vf will be 0:

d = (vf2 – vi2)/(2a) = [02 – (12.5 m/s)2]/[2(-9.81 m/s2)] = 7.96 m

Conclusion – If You Can Do 1-D, You Can Do 2-D

To conclude, 2-D kinematics questions are no different from 1-D kinematics questions, there is simply an extra step. Once we take the original velocity vector, and break it up into its components, you have transformed the 2-D question into 2 1-D questions. So if you can do 1-D kinematics, you can do 2-D too!