Projectile Motion Calculator with Steps | Range & Height

A projectile motion calculator finds how far, how high and how long an object flies after it is launched, using constant gravity and no air resistance.

A ball launched at 20.0 m/s at 30.0° on level ground (g = 9.81 m/s²) flies for 2.04 s, rises 5.10 m and lands 35.3 m away.

R and H in metres (m), v₀ in m/s, θ in degrees, g in m/s², T in seconds (s). These three forms are for level ground; from a height, use the full method in Section 6.

Rule of thumb: on level ground 45° gives the longest range, and two angles that add up to 90° (30° and 60°) land at the same spot. Double the speed and the range becomes four times longer.

Projectile motion is the curved path of an object that is thrown, kicked or fired and then moves under gravity alone. Its horizontal speed stays the same, while its vertical speed falls by 9.81 m/s every second. Together they trace a parabola.

1.1 How to calculate projectile motion in 5 steps

1. Convert to SI units: speed in m/s (km/h ÷ 3.6), heights in m, and choose g (9.81 m/s² or your syllabus value).
2. Split the launch velocity: v_x = v₀ cos θ and v_y = v₀ sin θ.
3. Find the time of flight: T = 2v_y/g on level ground, or T = (v_y + √(v_y² + 2gh))/g from a height h.
4. Find the maximum height: H = v_y²/(2g) above the launch point.
5. Find the range: R = v_x × T. For 20.0 m/s at 30.0°: 17.32 × 2.039 = 35.3 m.

1.2 Projectile motion formulas: which one to use

1.3 Horizontal and vertical projectile motion

Horizontal projectile motion is a launch at 0°, such as a ball rolled off a table or a stone thrown flat off a cliff. The fall time depends only on the height: t = √(2h/g). A stone thrown at 15.0 m/s from a 45.0 m cliff falls for 3.03 s and lands 45.4 m out. Vertical projectile motion is a throw straight up or down: a ball thrown up at 20.0 m/s rises 20.4 m.

1.4 Range and launch angle

On level ground the range R = v₀² sin 2θ / g is largest when sin 2θ = 1, that is at 45°. Angles the same distance either side of 45° give the same range. From a height the best angle is below 45°; from a cliff it can be far below. Air resistance also lowers it, to roughly 35 to 42° for sports balls.

1.5 Projectile motion with air resistance

Real balls feel a drag force that grows with the square of the speed. It shortens the range, lowers the peak and makes the path come down more steeply than it went up. There is no simple formula, so the Air Resistance tab solves it step by step: a 40.0 m/s baseball at 35.0° lands about 100 m away instead of 155 m.

1.6 What this calculator does

• Standard: enter speed, angle and launch height for every result, or solve backwards for the initial velocity (from range, maximum height or time of flight) or for the launch angle (from range, both answers).
• Horizontal launch: off a cliff or a table.
• Vertical: straight up or down, with a launch height.
• Air resistance: quadratic drag with mass, diameter and drag coefficient, compared with the vacuum path.
• Batch: many launches at once for a lab or an answer key.

Who uses it: CBSE and ICSE Class 11 and 12, JEE and NEET, IGCSE, A level and AP Physics 1 students; first-year university mechanics; coaches looking at throws and kicks; and anyone checking a launch in a game or an engineering sketch.

4.1 Two motions at once

Every projectile does two independent things. Sideways it moves at a steady v_x. Up and down it behaves exactly like a ball thrown straight up with speed v_y. Time is the only thing they share, which is why the time of flight links the range to the height.

4.2 Is your answer a sensible size?

A child throwing a ball manages about 10 to 15 m/s; a professional baseball pitch is about 40 m/s; a football kick about 25 m/s. At 20 m/s the longest possible vacuum range is about 41 m. If your range is thousands of metres, check for km/h left unconverted or radians typed as degrees.

4.3 What happens at the top

At the highest point the vertical velocity is zero, but the speed is not: the projectile is still moving sideways at v_x. The acceleration is still 9.81 m/s² downward. Exam questions test both facts.

4.4 Why 45° and when it is not

On level ground sin 2θ is largest at 45°, so that angle gives the longest range. From a height h the best angle is tan⁻¹(v₀ / √(v₀² + 2gh)), which is always below 45°. The calculator reports this best angle for your speed and height.

4.5 Complementary angles

On level ground 30° and 60° give the same range, as do 20° and 70°. The higher angle spends longer in the air and climbs higher. Solving for the angle from a range therefore gives two answers.

4.6 How the answer scales

Range and maximum height both grow with v₀², and time of flight with v₀. On the Moon, where g is about 1.62 m/s², the same throw goes about six times further and six times higher.

4.7 Landing speed and angle

Without air resistance the landing speed is √(v₀² + 2gh): the same as the launch speed on level ground and faster when landing lower down. On level ground the projectile lands at the same angle it left, mirrored below the horizontal.

4.8 When air resistance matters

For a heavy, slow object (a shot put, a stone) the vacuum answer is within a few per cent. For light, fast balls drag can cut the range by a third or more. Compare both in the Air Resistance tab.

▶ Press Calculate above to fill these five cards with your own answer.

This flowchart shows how a projectile motion calculator works, step by step, on the worked problem used across this page: a ball is launched at 20.0 m/s at 30.0° above level ground (g = 9.81 m/s²). Find its time of flight, maximum height and range.

From a height the steps are the same, but step 3 uses the quadratic formula. Sections 9, 10, 11 and 14 repeat this example in software and by hand.

1. Choose the type of launch. Pick Standard, Horizontal (off a cliff), Vertical, With Air Resistance or Batch.
2. Choose what to solve for. In the Standard tab, tap a chip: everything from speed and angle, or the initial velocity or launch angle from a range, height or time.
3. Type the launch speed and angle. The angle is measured up from the horizontal, in degrees unless you pick radians.
4. Add the initial height. Type how far the launch point is above the landing point. Leave it empty or 0 for level ground.
5. Pick units. Each box has its own unit list (m/s, km/h, mph, ft/s, m, ft). Choose how distances are shown in the settings row.
6. Choose g and significant figures. Pick 9.81, 9.8, 10 or 9.80665 m/s² to match your book, or the Moon or Mars.
7. Or load a sample problem. Thirteen samples cover level ground, 45°, a cliff, a shot put, solving backwards, a vertical throw, air resistance, mph and two traps. Nothing is calculated until you press Calculate.
8. Press Calculate. Or press Enter in any box. The answer appears with units, the other key results and two check badges.
9. Read the working. The GRESS table and numbered steps show the components, the time of flight, the height and the range, ready to copy.
10. Check the trajectory and charts, then download. The live diagram is drawn to scale and the four charts are built from your numbers. Download Doc saves the working as text; Download PDF prints it.

Excel can solve any projectile problem with ordinary formulas. Put one launch on each row, write the unit in every header, and remember that Excel's trig functions need radians. The worked numbers match Sections 7, 10 and 11: 20.0 m/s at 30.0° on level ground with g = 9.81 m/s².

9.1 The functions you need

9.2 Step by step

9.3 The awkward cases

• Trig in radians. =SIN(30) is −0.988. Always write SIN(RADIANS(30)).
• Square root of a negative. If the projectile cannot reach a higher landing level, SQRT returns #NUM!. That is the physics telling you it falls short.
• Solving for the angle. Use Data → What-If Analysis → Goal Seek on the range cell by changing the angle. Start once from 20° and once from 70° to find both answers.
• Charting the path. Make a column of t from 0 to T, then x = v_xt and y = h + v_yt − ½gt².

9.4 Why Excel might disagree with R, Python or this page

9.5 Charting it

1. Select the x and y columns for one launch.
2. Insert → Scatter → Scatter with Smooth Lines (never the Line chart type).
3. Format the axes with equal scales so the parabola keeps its true shape.
4. Label the axes "horizontal distance x (m)" and "height y (m)".

9.6 Errors you will hit

This base-R script solves the worked example (T = 2.04 s, H = 5.10 m, R = 35.3 m) for any launch height, checks the answer two ways, finds the best angle, and saves one 300 dpi trajectory plot.

10.1 The complete script

R, base only, saves projectile_trajectory.png

# ============================================================
# Projectile motion calculator in R: time of flight, maximum
# height, range (with or without a launch height), best angle,
# and one 300 dpi trajectory plot
# StatsUnlock - https://statsunlock.com/projectile-motion-calculator/
# Base R only. No packages to install.
# ============================================================

# ---- 1. Your values (SI units) -----------------------------
v0    <- 20.0   # launch speed, m/s
theta <- 30.0   # launch angle, degrees above horizontal
h     <- 0      # launch height above landing level, m
g     <- 9.81   # m/s^2 (9.80665 standard, 9.8 for CBSE/NCERT)

# ---- 2. Components (R trig functions use radians) ----------
th <- theta * pi / 180
vx <- v0 * cos(th)
vy <- v0 * sin(th)

# ---- 3. Time of flight, maximum height, range --------------
T_flight <- (vy + sqrt(vy^2 + 2 * g * h)) / g   # works for any h
t_top    <- max(vy, 0) / g
H        <- max(vy, 0)^2 / (2 * g)              # above the launch point
R        <- vx * T_flight

cat(sprintf("vx = %.4g m/s, vy = %.4g m/s\n", vx, vy))
cat(sprintf("Time of flight T = %.3g s\n", T_flight))
cat(sprintf("Maximum height H = %.3g m (at t = %.3g s)\n", H, t_top))
cat(sprintf("Range R = %.3g m\n", R))

# ---- 4. Checks ----------------------------------------------
if (h == 0) {
  R_short <- v0^2 * sin(2 * th) / g
  cat("Check R = v0^2 sin(2 theta)/g:", ifelse(abs(R - R_short) < 1e-9, "OK", "FAILED"), "\n")
}
v_land <- sqrt(vx^2 + (vy - g * T_flight)^2)
cat(sprintf("Landing speed = %.3g m/s (energy check: %.3g m/s)\n", v_land, sqrt(v0^2 + 2 * g * h)))

# ---- 5. Best angle for this speed and height ---------------
best <- atan(v0 / sqrt(v0^2 + 2 * g * h)) * 180 / pi
R_max <- v0 * sqrt(v0^2 + 2 * g * h) / g
cat(sprintf("Best angle = %.3g degrees, maximum range = %.3g m\n", best, R_max))

# ---- 6. One publication-ready figure: the trajectory -------
t <- seq(0, T_flight, length.out = 200)
x <- vx * t
y <- h + vy * t - 0.5 * g * t^2
png("projectile_trajectory.png", width = 6, height = 4, units = "in", res = 300)
par(mar = c(4.5, 4.5, 2.5, 1))
plot(x, y, type = "l", lwd = 3, col = "#2563eb", asp = 1,
     xlab = "horizontal distance x (m)", ylab = "height y (m)",
     main = sprintf("R = %.3g m, H = %.3g m, T = %.3g s", R, H, T_flight))
abline(h = 0, col = "#475569")
points(vx * t_top, h + H, pch = 19, col = "#9333ea")
invisible(dev.off())
cat("Saved projectile_trajectory.png\n")

Output when run: vx = 17.32 m/s, vy = 10 m/s; T = 2.04 s; H = 5.1 m at t = 1.02 s; R = 35.3 m; check OK; landing speed 20 m/s; best angle 45 degrees, maximum range 40.8 m.

10.2 What each part does

R traps: trig uses radians; round() goes half to even; -2^2 is −4; vectors recycle silently, so keep t one vector.

10.3 What the figure shows

A parabola from (0, 0) to (35.3, 0) m, peaking at 5.10 m above the point x = 17.7 m (purple dot). With asp = 1 one metre across equals one metre up, so the shallow 30° path looks as flat as it really is.

10.4 Common problems and fixes

• NaN for T: vy² + 2gh is negative, so the landing level is out of reach.
• Range looks wrong: check theta is in degrees in part 1 and converted once.
• No PNG: look in getwd().

10.5 Useful one-liners

The Python script repeats the worked example, solves for both launch angles that hit a target 30 m away (23.7° and 66.3°), and adds a baseball with air resistance using SciPy's solve_ivp. It saves the same 300 dpi trajectory.

Install once with pip install numpy scipy matplotlib.

11.1 The complete script

Python 3 with NumPy, SciPy and Matplotlib, saves projectile_trajectory.png

# ============================================================
# Projectile motion calculator in Python: range, maximum height,
# time of flight, solving for the angle, air resistance (RK45),
# and one 300 dpi trajectory plot
# StatsUnlock - https://statsunlock.com/projectile-motion-calculator/
# pip install numpy scipy matplotlib
# ============================================================
import math
import numpy as np
from scipy.integrate import solve_ivp
import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt

# ---- 1. Your values (SI units) -----------------------------
v0 = 20.0        # launch speed, m/s
theta = 30.0     # launch angle, degrees
h = 0.0          # launch height above landing level, m
g = 9.81         # m/s^2 (scipy.constants.g is 9.80665)

# ---- 2. No air resistance ----------------------------------
th = math.radians(theta)
vx, vy = v0 * math.cos(th), v0 * math.sin(th)
T = (vy + math.sqrt(vy**2 + 2 * g * h)) / g
H = max(vy, 0)**2 / (2 * g)
R = vx * T
print(f"vx = {vx:.4g} m/s, vy = {vy:.4g} m/s")
print(f"Time of flight T = {T:.3g} s, max height H = {H:.3g} m, range R = {R:.3g} m")
if h == 0:
    print("Check R = v0^2 sin(2 theta)/g:", "OK" if math.isclose(R, v0**2 * math.sin(2 * th) / g) else "FAILED")

# ---- 3. Launch angle needed to hit a target 30 m away ------
target = 30.0
A = g * target**2 / (2 * v0**2)
roots = np.roots([A, -target, A - h])            # quadratic in tan(theta)
angles = sorted(math.degrees(math.atan(r.real)) for r in roots if abs(r.imag) < 1e-12)
print("Angles to land 30 m away:", [round(a, 1) for a in angles], "degrees")

# ---- 4. With air resistance (quadratic drag) ---------------
m, d, Cd, rho = 0.145, 0.0737, 0.30, 1.225       # baseball
k = 0.5 * rho * Cd * math.pi * d**2 / 4 / m
def rhs(t, s):
    x, y, ux, uy = s
    sp = math.hypot(ux, uy)
    return [ux, uy, -k * sp * ux, -g - k * sp * uy]
def landed(t, s): return s[1] + 1.0              # launched 1.0 m above the ground
landed.terminal, landed.direction = True, -1
b0, bth = 40.0, math.radians(35.0)
sol = solve_ivp(rhs, [0, 30], [0, 0, b0 * math.cos(bth), b0 * math.sin(bth)],
                events=landed, rtol=1e-10, atol=1e-10, max_step=0.01)
print(f"Baseball 40 m/s at 35 deg with drag: range = {sol.y_events[0][0][0]:.4g} m")

# ---- 5. One publication-ready figure: the trajectory -------
t = np.linspace(0, T, 200)
x, y = vx * t, h + vy * t - 0.5 * g * t**2
fig, ax = plt.subplots(figsize=(6, 4))
ax.plot(x, y, color="#2563eb", lw=3)
ax.plot(vx * vy / g, h + H, "o", color="#9333ea")
ax.axhline(0, color="#475569", lw=1)
ax.set_aspect("equal")
ax.set_xlabel("horizontal distance x (m)")
ax.set_ylabel("height y (m)")
ax.set_title(f"R = {R:.3g} m, H = {H:.3g} m, T = {T:.3g} s")
fig.savefig("projectile_trajectory.png", dpi=300, bbox_inches="tight")
print("Saved projectile_trajectory.png")

Output when run: vx = 17.32 m/s, vy = 10 m/s; T = 2.04 s, H = 5.1 m, R = 35.3 m; check OK; angles to land 30 m away: 23.7 and 66.3 degrees; baseball with drag: range 100.2 m.

11.2 What each part does

Python traps: ** is power and ^ is XOR; scipy.constants.g is 9.80665, not 9.81; round(2.5) is 2.

11.3 What the figure shows

The same parabola as the R figure: 35.3 m long and 5.10 m high, with the peak marked. The drag result is printed rather than drawn: 100.2 m against 154.7 m in a vacuum.

11.4 Common problems and fixes

• ValueError: math domain error: the square root is of a negative, so the target level is out of reach.
• No angles printed: the target is beyond the maximum range.
• solve_ivp never lands: raise the time limit from 30 s for very long flights.

11.5 Useful one-liners

Run it in a terminal (python projectile.py), in Jupyter, or in Google Colab.

12.1 Range, height and time against launch angle (20.0 m/s, level ground, g = 9.81 m/s²)

Computed with the formulas in Section 6. Complementary angles (15° and 75°, 30° and 60°) share a range.

12.2 Gravity on other worlds

Standard gravity from the SI and NIST CODATA 2022; planetary surface values rounded from NASA planetary fact sheets.

12.3 Typical launch speeds

12.4 Drag data for the Air Resistance tab

Sea-level air density 1.225 kg/m³ at 15 °C (International Standard Atmosphere).

12.5 Unit conversions

The written method examiners expect, on the same example as Sections 7, 9, 10 and 11: a ball is launched at 20.0 m/s at 30.0° above level ground. Find the time of flight, maximum height and range (g = 9.81 m/s²).

14.1 The written method (GRESS)

1. Given. v₀ = 20.0 m/s, θ = 30.0°, h = 0, g = 9.81 m/s². Up and forward are positive.
2. Required. T (s), H (m), R (m).
3. Equations. v_x = v₀ cos θ, v_y = v₀ sin θ, T = 2v_y/g, H = v_y²/(2g), R = v_xT.
4. Substitute, with units. v_x = (20.0 m/s)(cos 30.0°) = 17.32 m/s; v_y = (20.0 m/s)(sin 30.0°) = 10.00 m/s.
5. Solve. T = 2(10.00)/9.81 = 2.039 s; H = 10.00²/19.62 = 5.097 m; R = 17.32 × 2.039 = 35.31 m.
6. Check. R = 20.0² sin 60° / 9.81 = 35.31 m. Units: (m/s)(s) = m. On level ground the landing speed is 20.0 m/s again.
7. Write the answer. T = 2.04 s, H = 5.10 m, R = 35.3 m (3 significant figures).

From a height: for the shot put (13.0 m/s, 38.0°, h = 2.10 m), v_y = 8.004 m/s, so T = (8.004 + √(8.004² + 2 × 9.81 × 2.10))/9.81 = 1.862 s and R = 10.24 × 1.862 = 19.07 m.

14.2 On a scientific calculator

The screen must show D (degrees). In R (radian) mode sin 30 gives −0.988 and every answer is wrong. Keep full figures until the last line.

14.3 Measure it yourself: a school lab practical

Measured ranges are about 2% short, typical of rolling losses. Paste "1.20, 0, 0.850" into the Batch tab to check each prediction.

14.4 Practice problems with answers

14.5 Mistakes that lose marks

• Calculator in radians mode, or degrees typed into a function that expects radians.
• Using R = v₀² sin 2θ / g when the launch and landing heights differ.
• Saying the velocity at the top is zero. Only v_y is zero; v_x carries on.
• Using sin for v_x and cos for v_y when the angle is measured from the horizontal.
• Taking the negative root of the time quadratic.
• Forgetting that only one of the two angles may make sense in context (a low throw under a roof).
• Giving the maximum height above the launch point when the question asks for the height above the ground.

Use the vacuum equations when:

• ✓ Gravity is the only force after launch (a dense, slow object over a short distance).
• ✓ g is constant: heights of a few kilometres or less.
• ✓ The ground is flat over the range: up to a few kilometres.

Use something else when:

• ✗ The object is light or fast (sports balls above about 20 m/s): use the Air Resistance tab.
• ✗ Launch and landing heights differ: use the general time-of-flight formula, not R = v₀² sin 2θ / g.
• ✗ Spin matters (a curving free kick, a topspin tennis shot): the Magnus force is beyond this tool.
• ✗ The range is hundreds of kilometres (long-range artillery, rockets): the Earth's curvature and changing g matter.
• ✗ The object has its own thrust (a rocket): it is not a projectile until the engine stops.

Real examples:

1. Exam questions in kinematics: Standard tab, with your board's g.
2. A stone thrown off a cliff or a ball rolled off a table: Horizontal tab.
3. A ball thrown straight up: Vertical tab.
4. The shot put: Standard tab with a launch height of about 2 m.
5. A basketball shot: Standard tab with a negative h, solving for the launch speed.
6. A baseball or golf drive: Air Resistance tab.

Assumptions

1. No air resistance (Standard, Horizontal, Vertical). Check that the object is heavy and slow. When drag matters, the vacuum answer overestimates range, height and time: a baseball at 40 m/s goes about 100 m, not 155 m.
2. Constant g. True to better than 0.1% below about 3 km altitude. At great heights g is smaller, so the formula underestimates the height reached.
3. Flat, non-rotating Earth. Over a few kilometres the error is tiny. Over long ranges the curving ground makes the formula underestimate the flight time and range.
4. A point mass. Spin and shape are ignored. Backspin adds lift and makes a real ball go further than the drag-only model; topspin makes it drop sooner.
5. Constant drag coefficient (Air Resistance tab). Real C_d changes with speed; for golf balls it drops sharply at high speed, so the model can underestimate the range of a fast drive.
6. Still air. A headwind shortens the range; a tailwind lengthens it.

Limitations

• No Magnus force (spin), wind or lift. Those need a full aerodynamic model.
• No rocket thrust: the object must be unpowered after launch.
• No relativistic effects (irrelevant below about 0.1c) and no orbital mechanics: for satellites, use orbital tools.
• The drag solution is numerical (fourth-order Runge-Kutta). It is checked by halving the time step, but it is not an exact formula.

What the answer is

A projectile motion calculator turns a launch speed, an angle and a height into a time of flight, a maximum height, a range and a landing velocity. For the worked example, 20.0 m/s at 30.0°, the ball flies for 2.04 s, rises 5.10 m and lands 35.3 m away.

What it means physically

The sideways motion is steady and the up-and-down motion is free fall; time links the two. That is why 45° is the best angle on level ground, why complementary angles share a range, and why launching from a height lowers the best angle.

Watch out for

Degrees vs radians, km/h vs m/s, the level-ground formula used from a height, and air resistance for light, fast balls. Say whether a height is above the launch point or above the ground.

Next steps

Find the launch speed you need with the velocity calculator, check the angle conversions with the angle calculator, and fit measured trajectories with polynomial regression (a parabola is a degree-2 fit). Come back to this projectile motion calculator for range, height, time of flight and air resistance, every step shown.

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