A video of this demonstration is available at this link.

This demonstration, as set up in the photograph above, is small enough to carry, or to wheel on a cart, to a remote location.

At left in the photograph is a projectile launcher, for which the projectile is one of the yellow plastic balls in the bag lying next to it. The launcher is aimed on a straight line to the center of the target hanging at right in the photograph. At the exit of the launcher is a photogate, which is connected to a control box, to the right of the laucher in the photograph. The control box is also connected to a drop box, from which the target is hanging. First making sure that the switch on the control box is set to “DISARM,” you load the launcher by placing the ball into its muzzle, then using the black rod (lying next to the bag of balls) to push the ball further into the muzzle. The launcher is spring loaded, and as you push the ball into it, you compress the spring. Three detents allow you to select low, medium and high muzzle velocity. As you push the ball in, you will hear a click at each detent. When you have pushed the ball in to the desired detent, remove the rod and set the switch on the control box to “ARM.” Then pull upwards and back on the white string hanging from the lever on top of the launcher barrel. The ball shoots out of the launcher, tripping the photogate as it exits, causing the target to drop. No matter which setting you choose, the ball always hits the falling target. At the lowest-speed setting, the ball meets the target just as it is about to hit the floor.

The launcher in the setup shown above is a PASCO ME-6800 short-range projectile launcher, whose three settings are perfect for the five-foot-long table and the 50-inch rod holding the drop box. Assuming that there is available a table that is five feet long and three feet high, with an edge that would allow the use of the clamps that hold the launcher and the drop box support, this apparatus could be fairly easily transported to a remote location. It would probably be possible to carry it, but if this were too awkward, the apparatus could be wheeled on a cart. Setting the demonstration up at the remote location would involve clamping the pieces as shown, making sure that the launcher is aimed at the target, then plugging the control box into an outlet.

The traditional setup uses a PASCO ME-6801 long-range projectile launcher set at one end of the main (eight-foot-long) demonstration table, with a 58-inch section of a much longer rod holding the drop box at the opposite end. Perhaps with a shorter rod, this would also be portable, but it would require an eight-foot-long table at the destination. With both setups, the three launcher settings yield similar results.

If you have a strong preference for one version or the other, please specify this when you request the demonstration. Depending on how many rolling tables are available, because of complaints about the target and its support structure interfering with the view of the chalkboard and screen, the smaller setup will probably become the default setup.

And now for the physics. . .

This demonstration is a variant of demonstrations 08.21 -- Simultaneous release, and 08.24 -- Ball over tunnel. Here, the ball starts with nonzero velocity in both the x and y directions, and is subject to acceleration only in the y direction, and it is shot towards a target that has zero velocity in the x direction, has zero initial velocity in the y direction, and is subject to the same acceleration in the y direction as the ball is.

The ball emerges from the launcher with a velocity v₀, whose angle with respect to the x direction is θ₀, the angle at which the launcher barrel is set. The components of this velocity are v_x0 = v₀ cos θ₀ and v_y0 = v₀ sin θ₀. Since there is no acceleration in the x direction, the horizontal component of the velocity is constant, and v_x = v₀ cos θ₀. In the y direction, the ball is accelerated downwards by gravity, so the y component is v_y = v₀ sin θ₀ - gt. The ball’s coordinates at any given time are x = v₀ cos θ₀t and y = v₀ sin θ₀t - (1/2)gt².

For the target, v_x = 0, and v_y = -gt. If we call the horizontal distance between the launcher and the target s, then the target’s height above the launcher is s tan θ₀, and its coordinates are x = s (always) and y = s tan θ₀ - (1/2)gt². For this demonstration to work, when the ball is at x = s, then its y coordinate must equal that of the target. That is, whenever the trajectory of the ball crosses the vertical line along which the target is falling, its y coordinate must equal that of the target. The time it takes for the ball to arrive at x = s is t = s/(v₀ cos θ₀). As noted above, the the y coordinates are y_ball = v₀ sin θ₀t - (1/2)gt², and y_target = s tan θ₀ - (1/2)gt². For these to equal, then v₀ sin θ₀t = s tan θ₀. Substituting s/(v₀ cos θ₀) for t gives v₀ sin θ₀(s/(v₀ cos θ₀)) = s tan θ₀, so when the x coordinates of the ball and target are equal, so are their y coordinates, and they collide. This is true as long as the magnitude of the velocity of the ball is great enough that the time it takes for the ball to reach x = s is less than or equal to the time it takes for the target to hit the floor.

As the equations show, if gravity did not act, the ball would fly along a straight line to the initial position of the target. Since both the ball and the target suffer the vertical acceleration of gravity, at any given time, the height of each is (1/2)gt² lower than what it would have been in the absence of gravity, so a projectile aimed along the straight line to the initial position of the target, if they are released simultaneously, always hits the target.

Another, equivalent, way to analyze this system is to consider the position vectors of the ball and the target, taking the launch point of the ball as the origin. The general equation for the position vector is r = r₀ + v₀t + (1/2)at². For the ball, the initial position is at the origin, and a = g, so r_ball = v_0ballt - (1/2)gt². That is, it is the resultant of the velocity vector and the acceleration due to gravity. For the target, r_target = r_0target - (1/2)gt². For a collision to occur, the ball and the target must be at the same position, or r_ball = r_target. This happens when v_0ballt = r_0target, or at t = r_0target/v_0ball, the time it would take the ball to reach the target along the straight line to the original position of the target if its velocity were constant.

Above, we noted that for the ball, x = v₀ cos θ₀t. If we rearrange this to get t in terms of x, and then substitute it into the equation for the y coordinate of the ball, we get y = v₀ sin θ₀(x/v₀ cos θ₀) - (1/2)gx²/(v₀ cos θ₀)², or y = tan θ₀ x - g/(2(v₀ cos θ₀)²) x². Since the initial velocity, initial angle and g are constants, we can express this as y = Bx - Ax², which is the equation of a parabola. Also, we can express v as the square root of the sum of the squares of its components, or v = √(v_x² + v_y²), from which we see that at any given time, v_x = cos θ, v_y = sin θ, and v_x/v_y = tan θ, where θ is the angle that v makes with the x direction.

References:

1) Resnick, Robert and Halliday, David. Physics, Part One, Third Edition (New York: John Wiley and Sons, 1977), pp. 55-59.
2) Young, Hugh D. and Geller, Robert M. Sears and Zemansky’s College Physics, 8^th edition (San Francisco: Addison Wesley, 2007), pp. 83-84.