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Physics Problem Solution: Projectile Motion

Well exam season is here. Hooray for all nighters and mental anguish!

So today I’m posting a comprehensive solution to a very common Physics 1 exam problem, pulled from a real university midterm.

Download the full PDF version of this Projectile Motion solution (with annotations) you can take with you

What you find when you dig deep, is that there’s actually a common theme among problems for each mechanics concept (in this case the kinematics of projectile motion). Physics professors are not always the most creative bunch when it comes to coming up with nasty unseen problems (and because physics is so hard for most students, they’d quickly send the average exam score down the toilet if they did).

So even though it feels like you’re seeing the material for the first time, there have actually been thousands and thousands and thousands of students who have come before you, who have seen that same exact problem, and freaked out in the same way you did.

Well I want to give you a leg up. Give you a glimpse into what you can expect of that fateful day, and ease the pain of the unknown. And hopefully improve your test scores in the process.

Let’s dive in.

Problem Statement:

A projectile is fired into the air from the edge of a 125-m high cliff at an angle of 30.2 deg above the horizontal. The projectile hits a target 455 m away from the base of the cliff. What is the initial speed of the projectile, v0?


Following our problem solving process, first things first, let’s restate the problem and identify exactly what we’re trying to find.

RESTATEMENT: Here we’re breaking down the known variables (initial height, angle of launch, and ending position of the projectile), as well as the unknowns we’re supposed to find (in this case the magnitude of the initial velocity).


GUESS: Next, I’m taking a stab at what I think the velocity could be just eyeballing the problem (asked myself how long it would take for something to fall 125 m and added a little to it). This establishes a baseline we can compare the final answer to.


Now that we’ve got our goal properly established in our minds, let’s get more specific by drawing a diagram and labeling the variables we’re going to need during the solution process.

DIAGRAM: Even though the problem gives a pretty good diagram, I’m drawing one myself to get a feel for how the projectile travels, establish my XY coordinate system, and label all the velocity and position variables.

Notice that I’m making it a point to establish symbolic labels for each piece of information given in the problem. This becomes especially important when it comes to doing our algebra correctly and catching potential mistakes. It’s also helpful when identifying which equations we might be able to use.

VARIABLES: To the right of my beautiful diagram is our list of variables based on what the problem gives us, plus gravity (assumed). I created theta, ho, d, and g so that I can solve symbolically, then substitute the numbers later.


With a clear, labeled drawing, and what we think are all of the variables we need, we can now move on to figuring out the relationships between them we might be able to use to solve for v0.

EQUATIONS: Because I’ve recognized this as a projectile problem, I know I need to use kinematics equations. And because the projectile is moving in both X and Y, I need to split the problem along those axes (because gravity acts along the Y axis and not X).

First I split the initial velocity into X and Y components using trig.


Beginning with the X-direction, I’m starting with the acceleration and integrating with respect to t (time), to get an equation for Vx (velocity in X), and then again to get an equation for x (position in X).

I want the position equation because I know the problem is giving me two different known positions (when it launches and when it lands), and I will have to use them to solve for v0.


I then repeat the same integration process to get an equation for y (position in Y).


Now I’m ready to start solving.

SOLUTION: Okay so here’s the tricky part of this problem – how to we solve for v0? Well we know we have two equations that use v0 (our x and y position equations). And we have information about two different positions in the equation – launch (x=0, y=125m) and landing (x=455m, y=0). So we need to relate x and y somehow using that information.

First option: use the initial position when t=0. I didn’t bother to test this out here because I realized that we already used this information to create our x and y position equations. Plugging in t=0 just gives us x=0 and y=ho, which gets us nowhere.

Second option: use the final position when t is some unknown value I’m calling td. This is how I solve the problem here.


I started with x first, plugging in t=td. At this time, we know x=d (our variable for the 455m). I then re-arrange to solve for td, knowing I can substitute that into my y equation next.

Next I plug td into my y equation. At t=td, we know that y equals 0 (the projectile hits the ground. I can then substitute for td the solution from my x equation above.


Now the only unknown in our equation is v0 (Hey that’s what we’re trying to find!).

All we have to do is use some algebra and re-arrange our equation to get to our solution for v0.


Plug and chug. And just double-check all of the units work out.


Final answer.

59 m/s is a bit higher than my initial guess of 10 m/s, but still within the same order of magnitude. If the answer turned out to be like 1,000 m/s, I would know something was wrong and go back and double-check my work.

So there you have it. All projectile problems tend to have the same typical structure, and are some of the most popular exam questions out there in mechanics.

If you learn this general structure, and repeat the process with a number of different problems, you’ll be more than prepared when one of these pops up on your midterm or final.


  • Kim bing August 14, 2015, 10:13 am


  • Tony October 28, 2015, 12:08 pm

    You’ve made two errors in your solution. You didn’t distribute the cos squared to Ho, and you didn’t distribute the 2 correctly in the denominator of the square root. Your final answer is correct, 59 m/s, but it doesn’t follow from the work you did.

    • Tom December 14, 2015, 10:47 am

      Thanks for the correction Tony – fixed.

  • Habtamu Asmerom October 30, 2015, 1:32 pm

    I need every question about physics

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