Analyze projectile motion by treating horizontal and vertical components independently

Learning goal: Explain why the horizontal and vertical components of a projectile's motion act independently when air resistance is negligible; solve for the range, time in the air, initial height, launch speed or final velocity of a horizontally launched projectile; and design and evaluate a prediction of where a ball rolling off a table will land, stating assumptions and sources of uncertainty.

One motion, two components. A projectile is an object moving through the air under gravity alone. Its motion looks curved, but the curve is what you get when two simple motions happen at once. Horizontally, nothing pushes or pulls on the projectile (air resistance is being ignored), so by Newton's first law its horizontal velocity stays constant. Vertically, gravity gives it a constant downward acceleration of 9.8 m/s², exactly as if it had no horizontal motion at all. The two components do not interact: the horizontal velocity has no effect on how long the object takes to fall, and falling has no effect on how fast it moves sideways. The only thing the components share is the clock. This is the whole content of the standard, and everything else is bookkeeping.

A classic demonstration makes the point: one ball is dropped from a table edge at the same instant a second ball is fired horizontally from the same height. They hit the floor together. The second ball travels farther sideways, but its vertical motion is identical to the first ball's. Filming both balls beside a grid shows the same drop in each equal time interval, while the fired ball moves the same horizontal distance each interval.

Setting up a horizontal launch. Choose an origin at the launch point, x positive in the direction of travel, and y positive upward (some people choose down positive; either works if you are consistent). A horizontally launched projectile has an initial velocity with a horizontal component v_x and a vertical component of zero. Then:

horizontal: x = v_x t (constant velocity, no acceleration);

vertical: y = −½gt² and v_y = −gt (starts from vertical rest, accelerates downward).

The time in the air is fixed by the vertical motion alone: an object launched from height h hits the ground when ½gt² = h, so t = √(2h/g). The range, the horizontal distance traveled, is then v_x times that time. Doubling the launch speed doubles the range but leaves the time unchanged; doubling the height multiplies the time by √2, not by 2.

A fully worked model. A ball rolls off a 0.90 m high table at 1.5 m/s. Time in the air: t = √(2 × 0.90 ÷ 9.8) = √0.1837 = 0.43 s. Range: x = 1.5 × 0.4286 = 0.64 m from the table's edge. Vertical velocity at impact: v_y = −9.8 × 0.4286 = −4.2 m/s. Final speed: √(1.5² + 4.2²) = √(2.25 + 17.64) = 4.5 m/s, directed at tan⁻¹(4.2 ÷ 1.5) = 70° below the horizontal. Notice that the horizontal velocity at impact is still 1.5 m/s, the value it had at launch. Positions every 0.10 s: (0.15, 0.85), (0.30, 0.70), (0.45, 0.46), (0.60, 0.12), measured from the floor; the x-values step evenly while the drops grow (0.05, 0.15, 0.25, 0.34 m), which is the signature of constant horizontal velocity paired with constant vertical acceleration.

Working backward. The same equations solve for any one unknown. If the range and height are known, the time comes from the height and the launch speed is range ÷ time. If the time in the air is known, the height is ½gt². If the launch speed and the range are known, the time is range ÷ speed and the height is ½gt². Solving each part for one quantity and passing it to the next part is the method; the quadratic formula is never needed for a horizontal launch, because the vertical motion starts from rest.

Common misconceptions. First, that a faster projectile stays up longer. It does not; only the height and g set the time. Second, that the projectile has some forward force keeping it going. It has none; it keeps its horizontal velocity because nothing acts to change it. Third, that the vertical acceleration is smaller because the ball is "mostly moving sideways." It is 9.8 m/s² downward at every instant regardless of the horizontal speed. Fourth, that the velocity at the top of an arc is zero. For a projectile launched at an angle, the vertical component is zero at the top but the horizontal component is unchanged, so the speed there is v_x, not zero.

Launches at an angle, briefly. Although problem solving in this course is limited to horizontal launches, the same independence applies to a launch at angle θ with speed v: v_x = v cos θ stays constant and v_y starts at v sin θ and decreases by 9.8 m/s each second. A launcher firing at 8.0 m/s at 30° above the horizontal on level ground has v_x = 6.9 m/s and v_y = 4.0 m/s; the ball rises for 4.0 ÷ 9.8 = 0.41 s to a height of 0.82 m, is in the air for 0.82 s and lands 6.9 × 0.816 = 5.7 m away. This is offered as an extension; the assessed skill is the horizontal case.

Investigation: predict where a rolling ball lands. Materials: a steel ball or marble, a short ramp clamped to a table so the ball leaves the table edge moving horizontally, two marks 0.50 m apart on the table, a stopwatch or a phone video, a meter stick, a plumb line, carbon paper or a sheet of paper with a target drawn on it, and tape. Steps: release the ball from a fixed height on the ramp and time it over the 0.50 m between the marks three times; speed = 0.50 m ÷ mean time. Measure the table height with the plumb line to find the floor point directly below the edge. Compute t = √(2h/g) and the predicted range, place the target there, and release the ball from the same ramp height. Record: the three times, the height, the predicted range and the actual landing distance for three trials. A result that supports the claim: landings that cluster around the target, no farther from it than the uncertainty in the predicted range (work that out from the spread of the three timings), with the same speed giving the same range. Sources of uncertainty and their effect: timing by hand is uncertain by about 0.1 s, which over a 0.3 s crossing is a large fraction and shifts the predicted speed and range up or down by the same fraction; the ball slows slightly while rolling on the table, so the speed at the edge is lower than the average over the marks and the ball lands short; and a plumb line off by 1 cm shifts every landing by 1 cm. Safety: keep the landing zone clear and catch the ball so it does not roll underfoot. What the model cannot show: air resistance and the ball's spin are ignored; the errors are small for a dense ball over half a meter but grow with speed and distance.

Worked example 1 — range of a horizontal throw from a cliff. A stone is thrown horizontally at 12 m/s from a cliff 45 m high. Time: t = √(2 × 45 ÷ 9.8) = √9.18 = 3.0 s (3.03). Range: 12 × 3.03 = 36 m (36.4). Vertical velocity at impact: −9.8 × 3.03 = −30 m/s (−29.7). Speed at impact: √(12² + 29.7²) = 32 m/s. If the stone had been thrown at 24 m/s instead, the time would still be 3.0 s and the range would double to 73 m.

Worked example 2 — launch speed from a landing point. A marble rolls off a 1.2 m high table and lands 0.80 m from the base. Time: √(2 × 1.2 ÷ 9.8) = √0.245 = 0.49 s (0.495). Launch speed: v_x = 0.80 ÷ 0.495 = 1.6 m/s. The height fixed the time; the range then fixed the speed. Reversing the order would be impossible, because the range alone says nothing about the time.

Worked example 3 — height from time in the air. A horizontally launched dart is in the air for 1.4 s. The launch height is ½ × 9.8 × 1.4² = 4.9 × 1.96 = 9.6 m. The launch speed was not needed and does not affect this answer; a faster dart from the same height would travel farther but land at the same moment.

Worked example 4 — two balls, one dropped and one fired. From a 2.0 m shelf, ball A is dropped and ball B is launched horizontally at 3.0 m/s at the same instant. Both take √(2 × 2.0 ÷ 9.8) = 0.64 s to reach the floor, so they land together. Ball A lands directly below; ball B lands 3.0 × 0.639 = 1.9 m out. At impact A moves at 6.3 m/s straight down; B has the same 6.3 m/s downward component plus 3.0 m/s sideways, a speed of √(6.3² + 3.0²) = 6.9 m/s.

Representation task: a strobe diagram from data. For the ball leaving the 0.90 m table at 1.5 m/s, make a table of t, x and y at 0.10 s intervals (0, 0.10, 0.20, 0.30, 0.40 s and the landing time 0.43 s). Plot the points on graph paper with the same scale on both axes. Draw a vertical arrow at each point for v_y (length proportional to 9.8t) and a horizontal arrow for v_x (always the same length). Write two sentences, one explaining why the horizontal spacing is even and one explaining why the vertical drops increase.

Independent performance: hit the target. Carry out the rolling-ball investigation from the lesson, or the accelerated-motion version: release a low-friction cart on a ramp, collect position vs. time data, find its acceleration from the graph (tangent slopes plotted as a velocity vs. time graph, or a fit to x = ½at²) and use it to predict the cart's speed at the ramp's end. Then predict the landing point for a launch from the table edge and test it with a target. Success criteria: the launch speed is measured, not guessed, with its method and uncertainty stated; the time in the air is computed from the height only; the predicted range is stated before the trial; the landing distance is compared with the prediction as a percentage; and at least two sources of uncertainty are named with the direction in which each one shifts the landing.