Projectile Motion

A projectile is shot at an initial velocity v₀ at an angle q above the horizon. How far will it go? How high will it rise? What is the shape of its trajectory?

Denote the projectile's range R and the height of its trajectory h. Let us choose the origin of the coordinate system at the launch point, the x axis in the horizontal direction and the y axis in the vertical direction pointing up. Then the components of v₀, the initial velocity vector, are


	v_0x = v₀ cos q v_0y = v₀ sin q	(1)

The entire motion is described by the kinematics vector equations that prescribe the variation with time of position vector r and velocity vector v for a given acceleration vector a:


	dr/dt = v dv/dt = a

Each vector equation is actually a set of two independent equations (this is a two dimensional motion) for the x and y components of the corresponding vectors. That is, these equations are simply shorthand notation for


	dx/dt = v_x dv_x/dt = a_x
	dy/dt = v_y dv_y/dt = a_y

So in reality we are dealing with two sets of independent equations for the x and y directions and can envision the problem as a superposition of two completely independent motions that together are equivalent to the projectile motion.

Let us consider two other objects. The first one moves in the horizontal direction with the constant velocity v_0x. All the factors that govern the motion of this object and of the horizontal component of our projectile motion are identical. These two motions cannot be distinguished from each other. Therefore, if we solve for the motion of this imaginary object we will have solved the x component of the projectile motion. The second object we consider is thrown straight up with velocity v_0y. All the factors that govern the motion of this object and of the vertical component of our projectile motion are identical. Again, if we solve for the motion of this imaginary object we will have solved the y component of the projectile motion. These two imaginary objects will become our projectile when we link their motions with a common time because in the projectile case, both the horizontal and vertical motions start at the same place at the same time. Our projectile is equivalent to an object moving up and down, and also sliding in the horizontal direction while executing this vertical motion.

We can easily find the total duration of the vertical component of the motion; you may wish to revisit the relevant note on up-and-down motion to refresh your memory and see how that's done. Once we know how long the projectile stayed in the air, finding the range from the horizontal motion is simple. Let's put these ideas together.

The Horizontal Component
Neglecting air resistance, the velocity x component does not change throughout the entire motion. Therefore, v_x = v_0x and


	x = v_0xt	(2)

That's all for the horizontal motion.

The Vertical Component
The y component of the projectile's motion is subject to the gravitational acceleration which points downward. So v_y as a function of time is


	v_y = v_0y - g t	(3)

for our choice of axes. You may wish to revisit the note on the choice of signs for velocity and acceleration. The y motion can be envisioned as an object thrown straight up with velocity v_0y. Such an object will rise until its velocity is zero, so from the last equation the time to reach maximum height is


	t_max = v_0y/g	(4)

The object will then fall to the ground and the downward trip is the mirror image of the trip up; you may wish to revisit the relevant note. Therefore, the total time the object stays in the air is


	t_tot = 2t_max = 2v_0y/g	(5)

Putting it Together
That's it. We have derived the total duration of the projectile's motion. To find the range R, all that is left is to plug t_tot into equation (2) which gives us the distance covered at any time t. The result is


R	= v_0xt_tot
	= 2v_0xv_0y/g
	= 2(v₀²/g) sin q cos q	(6)

where in the last step we inserted v_0xand v_0y from equation (1).

Why the funny dependence on angle? From the first step in equation (6), the range is the product of the horizontal velocity component and the total duration of the motion. For a given launch speed v₀, large v_0x are obtained at small launch angeles. But then the duration of the flight is short because v_0y is small. Conversely, a steep launch angle will maximize the duration of the motion by giving a large v_0y, but then v_0x is small and the object simply goes up and down without getting very far. The competition between these two opposing effects is the reason for the q dependence shown in our result. This result can be simplified when we recall the identity from trigonometry sin2q = 2sinq cosq. With this identity, the range becomes simply


	R = (v₀²/g) sin2q	(7)

The maximum value of the sine function is 1, obtained when its argument is p/2 (90°). Therefore, for a given v₀, the maximal range is v₀²/g, obtained for a launch angle q = 45°. The function sinq is symmetric about its peak at 90°, therefore the function sin2q is symmetric about 45°. For example, a projectile launched at 30° will have the same range as one launched at 60°.

The Trajectory Height
An object thrown upward with velocity v_0y will rise to a height v_0y²/2g; you may wish to revisit the note on up-and-down motion to verify this result. So this is the height of the projectile's trajectory


	h = v_0y²/2g = (v₀²/2g) sin²q

The Trajectory Shape
At any time t, the y coordinate obeys


	y = v_0yt - ½ g t²	(7)

for our choice of y axis (check again the note on the choice of signs). From equation (2) we can eliminate time in terms of horizontal distance, t = x/v_0x. With this replacement, equation (7) is transformed into a relation for y as a function of x


	y = Ax - Bx²

where the coefficients A and B are


	A = tanq B = ½ g/(v₀ cosq)²

Verify these results for yourself, performing the replacement t = x/v_0x in equation (7).

The function we just derived shows that y has a quadratic dependence on x, namely, the trajectory is a parabola. y vanishes at x = 0 and x = A/B. Can you figure out the meaning of these two zeroes of the function?

[Lecture Notes] [PHY231] [Physics Dept] [UK]