Projectile motion

On a field you release a ball at an angle to the horizontal. For a moment it rises, then it falls and lands farther than the launch point. That is projectile motion: motion in a plane under constant g, with initial speed at angle α.

The body gets speed v₀ at angle α to the horizontal. In this model there is no air drag, so the path is a parabola. From that one picture you get range, maximum height, flight time, and the state at any instant.

First split v₀ into components:

v_x = v₀·cos α (constant)

v_y0 = v₀·sin α

Acceleration: a_x = 0, a_y = -g. With launch at height h₀ above the landing level: x(t) = (v₀·cos α)·t and y(t) = h₀ + (v₀·sin α)·t - ½·g·t². Measure α from the horizontal, not from the vertical.

On flat ground, h₀ = 0, landing y = 0 gives the flight time:

T = 2·v_y0/g = 2·v₀·sin α / g

Range and maximum height above the start have the canonical form:

Z = v₀²·sin(2α) / g

H_max = v₀²·sin²α / (2g)

These are the same ideas as in a vertical throw; only the vertical component does the work. sin(2α) has a maximum of 1 at α = 45°, so on flat ground Z_max = v₀²/g. First numeric intuition: with v₀ = 20 m/s, α = 45°, and g = 9.81 m/s² you get Z ≈ 40.8 m and H_max ≈ 10.2 m. That is a quick check in the range calculator and in full projectile motion.

Second intuition: the same v₀ = 20 m/s, but α = 30°. Then H_max = v₀²·sin²30°/(2g) = 400·(1/4)/19.62 ≈ 5.10 m. Flat-ground range matches the 60° case, because sin(60°) = sin(120°), yet the path is lower and flatter than at 45°. Compute the height in the H_max calculator.

Units: v₀ in m/s, α in degrees (calculators usually use degrees), g in m/s², Z and H_max in meters. If your speed is in km/h, divide by 3.6 before substituting. 72 km/h is 20 m/s.

When h₀ > 0 (tower, roof), the classic formula Z = v₀²·sin(2α)/g no longer holds, because flight time is no longer 2·v_y0/g. You must solve the quadratic y(t) = 0 for the positive root t, then Z = v_x·t. Height above the ground is h₀ + v_y0²/(2g) if v_y0 > 0. That is why the full model exists, not only a “mainly Z” page.

Model limits help you pick the tool. α → 0° with h₀ > 0 becomes a horizontal throw: t → √(2h₀/g). α → 90° with h₀ = 0 is a vertical throw: Z → 0, H_max → v₀²/(2g). Choosing the model on purpose saves half a page of algebraic mistakes.

Component intuition: the horizontal part moves uniformly and builds Z; the vertical part sets time and H_max. On flat ground, 45° is the best compromise between long hang time and large v_x. From a tower the optimum angle is usually less than 45°, because part of the range is “bought” by height h₀. So do not copy a field angle onto a roof launch automatically.

A common mistake sounds like this: “I throw from a roof, so I use Z = v₀²·sin(2α)/g and I have the range.” The formula looks familiar, but it assumes a return to launch level. From a roof the flight lasts longer, so real Z is larger, and you need the root of y(t) = 0. Other traps: measuring α from the vertical instead of the horizontal, and believing that at 60° flat-ground range is larger than at 30°.

Energy with no drag checks consistency: ½·m·v₀² + m·g·h₀ = ½·m·v² + m·g·y. At the apex v_y = 0, so v_x and height remain. You cannot extract range from the balance alone: you need flight time from kinematics.

Name three heights in a tower problem. h₀ is launch above the ground. H_rel = v_y0²/(2g) is the rise above the start. H_abs = h₀ + H_rel is the apex above the ground. Range Z is measured to y = 0 on the ground, so flight time mixes those levels.

The path equation y(x) comes from eliminating time: y = h₀ + x·tan α - (g·x²)/(2·v₀²·cos²α). It helps when you know an obstacle at distance x and ask whether the path clears it. You still assume no drag and constant g.

Before calculating, list v₀, α, h₀, g. Find v_x and v_y0. If h₀ = 0, you may use the ready Z, H_max, and T formulas. If h₀ ≠ 0, set y(t) = 0 and take the physical root t > 0. Finally check limits: at α = 45°, the order of Z is about v₀²/g.

Quick flat-ground sanity check: at α = 45° and v₀ = 20 m/s the range should be near 40.8 m. If you get a few meters or a few kilometers, you almost certainly mixed speed units or entered the angle in radians where the calculator expects degrees.

When the problem asks only for H_max, you do not need the full range. The vertical component and the vertical-throw-style formula are enough. Conversely, an energy balance alone will not tell you where the ball lands horizontally. For range Z you always come back through flight time and constant v_x.

The model breaks down with strong drag, ball spin, or crosswind. Then the no-drag parabola is more of an upper estimate of range. For field problems at modest v₀ it is good enough. The next topic in the series is the inclined plane.

Before you trust the answer, compare it with a second path: same v₀, different angle. If 45° is not the longest range on flat ground, or if 30° and 60° give different Z values, go back to the components and the units. Comparing two angles catches a mistake faster than substituting into the same formula a third time.