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Projectile Motion Equations for Intro Physics

Updated on October 6, 2016

A particle moves in a vertical plane with an initial velocity vo, but its acceleration is always the free-fall acceleration g= 9.8m/s2, which is directed downward. Such a particle is called a projectile, and it's motion is called projectile motion. Many sports deal with this sort of motion. Let's take a look at how to solve these problems. Firstly note, that we assume air resistance is negligible aka is not considered into the problem.

The projectile is launched with an initial velocity vo:


The components vox and voy can be found if we know the angle Θo between vo and the positive x-direction:

vox=vocosΘo and voy=vosinΘo

During it's two-dimensional motion, the projectile position vector, r, and velocity, v, changes throughout but it's acceleration vector, a, is constant and always directed vertically downward. The projectile has no horizontal acceleration. And neither the horizontal nor vertical motion affect the other- they are independent of each other.

We break up a projectile problem into two simpler problems: horizontal (acceleration= 0m/s2) and vertical (constant downward acceleration= 9.8m/s2)

Now that we have that information we can start analyzing projectile motion.

Horizontal Motion

Again: There is no horizontal acceleration, so a=0m/s2. We can start with equation 1.

(1) Δx=v0t+ (1/2)at2

Since we know a=0m/s2, we can see that equation 1 changes to 1a.

(1a) Δx=v0t

(1a) (x-x0)= v0t

For the horizontal component, we also know that vox=vocosΘo. Subbing that into 1a, we get another equation we can now plug numbers into, 1b.

(1b) Δx= (v0cosΘ0)t

(1b) (x-x0)= (v0cosΘ0)t

Vertical Motion

For the vertical motion we know that the acceleration is free-fall acceleration directed downward. This means that a=-g throughout the projectile motion, for the vertical component. (a= -9.8m/s2). We can also use the same equation we used for the horizontal component of motion, but in terms of y, since are dealing with the vertical motion. See equation 2 and 2a.

(2) Δy=v0t+ (1/2)at2

(2a) (y-y0)=v0t+ (1/2)at2

For the vertical component, we know that voy=vosinΘo, therefore we get the equation in 2b and 2c.

(2b) Δy=(v0sinΘ0)t+ (1/2)(-g)t2

(2c) (y-y0)=(v0sinΘ0)t - (1/2)gt2

The equations

(3) vf=vo+ at


(4) vf2=vo2+ 2aΔx

will also be useful equations when dealing with the vertical motion of a projectile problem and can be transformed. When plugging in what we know for the acceleration and velocity for vertical motion, as discussed earlier, we can get the equations in 3a and 4a.

(3a) vy= v0sinΘ0 - gt


(4a) vy2= (v0sinΘ0)2- 2gΔy

The velocity is directed upward initially and the magnitude of the velocity steadily decreases until it reaches 0. Once it reaches 0m/s, it has reached the maximum height.

The Trajectory

If we solve for t in equation 1 and substitute that into equation 2, with a little rearranging we can get equation 5, which is the path of the projectile, aka it's trajectory.

(5) y= (tanΘ0)x- (gx2/[2(v0cosΘ0)2]

The Horizontal Range

The horizontal range is the distance the projectile traveled horizontally. In equation 1, instead of having (x-xo) we will have R for range, as shown in equation 6. And since we are looking at the horizontal range (distance along x), in equation 2, we know that (y-yo=0), as shown in equation 7.

(6) R= (v0cosΘ0)t


(7) 0= (v0sinΘ0)- (1/2)gt2

The next thing we can do is eliminate t from bother equations 6 and 7, producing equation 8.

(8) R= ([2v02]/ g)sinΘ0 cosΘ0

Next we have to use the identity0= 2sinΘ0cosΘ0 and from that we obtain equation 9, representing the range.

(9) R= ([v02]/g) sin 2Θ0

*This equation is for the range of a projectile when the initial launch height is equivalent to the final launch height.


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