After completing this section you should be able to resolve a velocity vector into horizontal and vertical components and solve each separately by using standard equations of motion.
Motion in two dimensions
Up until this point we have only been considering motion along a straight line ( i.e one dimensional motion). We will now consider objects that are moving in a two dimensional plane. To visualise this imagine a small dot moving on a thin flat sheet of paper. The dot can move in any direction on the surface of the paper but it cannot move away from the surface. i.e. Its motion is restricted to two dimensions (it can move in any direction along the length or width of the papers sheet (or in a direction that combines movement in both e.g. diagonally)).
In this section we will consider motion in a vertical plane (imagine holding the sheet of paper vertically rather than horizontally! ).
(Note: There is nothing "restricting" the motion to two dimensions in these examples it is just the natural motion of the object's being considered.)
A projectile is an object that is launched into the air at an angle to the surface. After leaving the ground the only forces that act on the object are gravity and a drag force (due to air resistance as it move through the air). At this introductory level we will ignore air resistance and consider the only force acting to be gravity This results in the object travelling along a curved path in a vertical plane until it returns to the ground.
The key point to understand in projectile motion is that the vertical and horizontal motion of the object are completely independent and therefore we can analyse the vertical and horizontal motion separately.
To visualise the horizontal motion imagine looking down from directly above the path so that you were unable to "see" any evidence of the object's vertical motion. In this case the object would just appear to move along a straight line with constant velocity between two points.
We can use equation (2) of the "suvat"" equations (s = ut + 1/2at2) to determine its position along the x axis at any given time
This simplifies to s = ut (because a = 0)
Finally we will use x instead of s and ux instead of u to help remind us that this is the equation is for motion along the x axis.
x = uxt (i)
Imagine looking edge on to the plane of motion from the side, so that you are unable to "see" any evidence of the object's horizontal motion. From this view point it would appear as if the object had been launched vertically into the air and then fell back down due to gravity.
Therefore we can use the variations of suvat equations (1) and (2) derived in the previous section to describe the vertical component of the projectile's motion
y = uyt - 1/2gt2 (ii)
vy = uy - gt (iii)
As in the previous section we have used y instead of s to remind us that the displacement is along the y axis and used g instead of a to remind us that the acceleration is acceleration due to gravity etc.
In order to use the above equations we need to know both the object's initial velocity along the x axis (ux) and its initial velocity
along the y axis (uy).
In projectile problems we are often given the object's initial velocity (u)with the direction specified by the angle (Θ) between the velocity vector and the surface it is being launched from. From this information the horizontal and vertical components of the velocity (i.e. the velocity along the x and y axis respectively (ux &uy)) can be determined using simple trigonometry.
If an object is launched into the air with an initial velocity u in a direction at an angle to the ground of Θ: