Projectile motion refers to the motion of a particle that is imparted with an initial velocity but is subsequently subjected to no forces besides that of gravity.
This includes problems in which a particle is hurled at an angle between 0 and 90 degrees to the horizontal, with the horizontal usually being the ground. For convenience, these projectiles are assumed to travel in the (x, y) plane, with x representing horizontal displacement and y vertical displacement.
The path taken by a projectile is referred to as its trajectory. (Note that the common link in "projectile" and "trajectory" is the syllable "-ject," the Latin word for "throw." To eject someone is literally to throw him out.) The point of origin of the projectile in problems in which you need to calculate the trajectory is usually assumed to be (0, 0) for simplicity unless otherwise stated.
The trajectory of a projectile is a parabola (or at least traces a portion of a parabola) if the particle is launched in such a way that is has a nonzero horizontal motion component, and there is no air resistance to affect the particle.
The Kinematic Equations
The variables of interest in the motion of a particle are its position coordinates x and y, its velocity v, and its acceleration a, all in relation to a given elapsed time t since the start of the problem (when the particle is launched or released). Note that the omission of mass (m) implies that gravity on Earth acts independently of this quantity.
Note also that these equations ignore the role of air resistance, which creates a drag force opposing motion in real-life Earth situations. This factor is introduced in higher-level mechanics courses.
Variables given a subscript "0" refer to the value of that quantity at time t = 0 and are constants; often, this value is 0 thanks to the coordinate system chosen, and the equation becomes that much simpler. Acceleration is treated as constant in these problems (and is in the y-direction and equal to -g, or –9.8 m/s2, the acceleration owing to gravity near Earth's surface).
x = x0 + vx t
vx is the constant x-velocity. .
- y = y0 + [(v0y + vy)/2]t
- vy = v0y – gt
- y = y0 + v0yt –(1/2)gt2
- vy2 = v0y2 – 2g(y – y0)
Examples of Projectile Motion
The key to being able to solve problems that include trajectory calculations is knowing that the horizontal (x) and vertical (y) components of motion can be analyzed separately, as shown above, and their respective contributions to overall motion neatly summed at the end of the problem.
Projectile motion problems count as free-fall problems because, no matter how things look right after time t = 0, the only force acting on the moving object is gravity.
- Be aware that because gravity acts downward, and this is taken to be the negative y-direction, the value of acceleration is -g in these equations and problems.
1. The fastest pitchers in baseball can throw a ball at just over 100 miles an hour, or 45 m/s. If a ball is thrown vertically upward at this speed, how high will it get and how long will it take to return to the point at which it was released?
Here vy0 = 45 m/s, -g = –9.8 m/s, and the quantities of interest are the ultimate height, or y, and the total time back to Earth. Total time is a two-part calculation: time up to y, and time back down to y0 = 0. For the first part of the problem, vy, when the ball reaches its peak height, is 0.
Start by using the equation vy2 = v0y2 – 2g(y – y0) and plugging in the values you have:
0 = (45)2 – (2)(9.8)(y – 0) = 2,025 – 19.6y
y = 103.3 m
The equation vy = v0y – gt shows that the time t this takes is (45/9.8) = 4.6 seconds. To get total time, add this value to the time it takes for the ball to fall freely to its starting point. This is given by y = y0 + v0yt – (1/2)gt2 , where now, because the ball is still at the instant before it begins to plummet, v0y = 0.
Solving (103.3) = (1/2)gt2 for t gives t = 4.59 seconds.
Thus the total time is 4.59 + 4.59 = 9.18 seconds. The perhaps-surprising result that each "leg" of the trip, up and down, took the same time underscores the fact that gravity is the only force in play here.
2. The range equation: When a projectile is launched at a velocity v0 and an angle θ from the horizontal, it has initial horizontal and vertical components of velocity v0x = v0(cos θ) and v0y = v0(sin θ).
Because vy = v0y – gt, and vy = 0 when the projectile reaches its maximum height, the time to maximum height is given by t = v0y/g. Because of symmetry, the time it will take to return to the ground (or y = y0) is simply 2t = 2v0y/g.
Finally, combining these with the relationship x = v0xt, the horizontal distance traveled given a launch angle θ is
R (range) = 2(v02sin θ ⋅ cos θ/g) = v02(sin2θ)/g
(The final step comes from the trigonometric identity 2 sinθ ⋅ cosθ = sin 2θ.)
Since sin2θ is at its maximum value of 1 when θ = 45 degrees, using this angle maximizes the horizontal distance for a given velocity at
R = v02/g.