Imagine a ball rolling on a horizontal surface that is illuminated by a stroboscopic light. Figure (a) shows the position of the ball at even intervals of time along a dotted path. Case 1 is illustrated in positions 1 through 3; the magnitude and direction of the velocity do not change (the pictures are evenly spaced and in a straight line), and therefore, there is no acceleration. Case 2 is indicated for positions 3 through 5; the ball has constant speed but changing direction, and therefore, an acceleration exists. Figure (b) illustrates the subtraction of v 3 and v 4 and the resulting acceleration toward the center of the arc. Case 3 occurs from positions 5 to 7; the direction of the velocity is constant, but the magnitude changes. The acceleration for this portion of the path is along the direction of motion. The ball curves from position 7 to 9, showing case 4; the velocity changes both direction and magnitude. In this case, the acceleration is directed nearly upward between 7 and 8 and has a component toward the center of the arc due to the change in direction of the velocity and a component along the path due to the change in the magnitude of the velocity.
(a) Path of a ball on a table. (b) Acceleration between points 3 and 4.
Anyone who has observed a tossed object—for example, a baseball in flight—has observed projectile motion. To analyze this common type of motion, three basic assumptions are made: (1) acceleration due to gravity is constant and directed downward, (2) the effect of air resistance is negligible, and (3) the surface of the earth is a stationary plane (that is, the curvature of the earth's surface and the rotation of the earth are negligible).
To analyze the motion, separate the two‐dimensional motion into vertical and horizontal components. Vertically, the object undergoes constant acceleration due to gravity. Horizontally, the object experiences no acceleration and, therefore, maintains a constant velocity. This velocity is illustrated in Figure where the velocity components change in the y direction; however, they are all of the same length in the x direction (constant). Note that the velocity vector changes with time due to the fact that the vertical component is changing.
In this example, the particle leaves the origin with an initial velocity ( v o ), up at an angle of θ o. The original x and y components of the velocity are given by v x0 = v o and v y0 = v o sin θ o .
With the motions separated into components, the quantities in the x and y directions can be analyzed with the one‐dimensional motion equations subscripted for each direction: for the horizontal direction, v x = v x0 and x = v x0 t; for vertical direction, v y = v y0 − gt and y = v y0 − (1/2) gt 2, where x and y represent distances in the horizontal and vertical directions, respectively, and the acceleration due to gravity ( g) is 9.8 m/s 2. (The negative sign is already incorporated into the equations.) If the object is fired down at an angle, the y component of the initial velocity is negative. The speed of the projectile at any instant can be calculated from the components at that time from the Pythagorean theorem, and the direction can be found from the inverse tangent on the ratios of the components:
Other information is useful in solving projectile problems. Consider the example shown in Figure where the projectile is fired up at an angle from ground level and returns to the same level. The time for the projectile to reach the ground from its highest point is equal to the time of fall for a freely falling object that falls straight down from the same height. This equality of time is because the horizontal component of the initial velocity of the projectile affects how far the projectile travels horizontally but not the time of flight. Projectile paths are parabolic and, therefore, symmetric. Also for this case, the object reaches the top of its rise in half of the total time (T) of flight. At the top of the rise, the vertical velocity is zero. (The acceleration is always g, even at the top of the flight.) These facts can be used to derive the range of the projectile, or the distance traveled horizontally. At maximum height, v y = 0 and t = T/2; therefore, the velocity equation in the vertical direction becomes 0 = v o sin θ − g T/2 or solving for T, T = (2 v 0 sin θ)/ g.
Substitution into the horizontal distance equation yields R = ( v o cos θ) T. Substitute T in the range equation and use the trigonometry identity sin 2θ = 2 sin θ cos θ to obtain an expression for the range in terms of the initial speed and angle of motion, R = ( v o 2/ g) sin 2θ. As indicated by this expression, the maximum range occurs when θ = 45 degrees because, at this value of θ, sin 2θ has its maximum value of 1. Figure sketches the trajectories of projectiles thrown with the same initial speed at differing angles of inclination.
Range of projectiles launched at different angles.
For uniform motion of an object in a horizontal circle of radius (R), the constant speed is given by v = 2π R/ T, which is the distance of one revolution divided by the time for one revolution. The time for one revolution (T) is defined as period. During one rotation, the head of the velocity vector traces a circle of circumference 2π v in one period; thus, the magnitude of the acceleration is a = 2π v/ T. Combine these two equations to obtain two additional relationships in other variables: a = v 2/ R and a = (4π 2/ T 2) R.
The displacement vector is directed out from the center of the circle of motion. The velocity vector is tangent to the path. The acceleration vector directed to the center of the circle is called centripetal acceleration. Figure shows the displacement, velocity, and acceleration vectors at different positions as the mass travels in a circle on a frictionless horizontal plane.
Uniform circular motion.