Projectile Motion Simulator

parabolic trajectory with live animation

Starting values — Projectile Motion Simulator

Launch angle (θ)	45
Launch height (h)	0

Change any control above and the results recompute from these.

Loading interactive simulation...

Forty-five degrees wins only on flat ground 🖖

Projectile motion is the kinematic analysis of an object launched into a gravitational field, assuming constant acceleration g downwards. Neglecting air resistance, the horizontal velocity remains constant, while the vertical motion behaves as a uniformly accelerated system, producing a parabolic path. The maximum range is achieved at an angle of 45 degrees, where the horizontal and vertical velocity components are balanced.

Two motions stacked into one 🖖

A projectile is really doing two independent things at once: drifting sideways at a steady pace while falling under gravity. Neither influences the other, which leads to a famous result — a bullet fired horizontally and one simply dropped from the same height hit the ground at the exact same moment. Change the launch speed here and watch the sideways reach grow while the time in the air stays tied only to the vertical motion.

Above ground, 45° stops winning 🖖

The famous 45° rule for maximum range only holds when you launch and land at the same height. Add a launch height h and the best angle drops below 45°, following θ = arctan(v / √(v² + 2gh)). Throw a ball off a cliff and aiming a little flatter than 45° actually sends it farther — raise the height slider and hunt for the true optimum.

PROJECTILES — THE ANGLE DECIDES EVERYTHING EXCEPT WHAT THE SPEED DECIDES

Which Launch Case Are You Solving?

Horizontal and vertical motion are independent: constant speed sideways, constant acceleration downwards. Everything else follows from that split. On level ground the range is R = v²sin(2θ)/g, which peaks at 45° and pairs up angles either side of it. Launch from a height and that symmetry breaks, and the best angle drops below 45°.

45° on level ground — the maximum range R = v²/g at θ = 45°

Complementary angles — two ways to hit the same spot R = v²sin(2θ)/g

Straight up — all height, no range h = v²/2g, R = 0

Launched from a height — the symmetry breaks h > 0 ⇒ θ_opt < 45°

45° on level ground — the maximum range

What you know: Launch and landing are at the same height, and you want the furthest throw. sin(2θ) is largest when 2θ = 90°.

Range: R = v²/g at θ = 45°

Worked example: v = 20 m/s at 45° → R = v²/g = 400/9.81 = 40.8 m, the longest this speed can reach

Open this case: 45° max range

45° on level ground — the maximum range. At 45° the hang time and the horizontal speed are balanced, and the range peaks. Launch and landing are at the same height, and you want the furthest throw. sin(2θ) is largest when 2θ = 90°. — At 45° the hang time and the horizontal speed are balanced, and the range peaks.

Complementary angles — two ways to hit the same spot

What you know: Any angle θ and its complement 90° − θ give the same range on level ground, because sin(2θ) = sin(180° − 2θ).

Range: R = v²sin(2θ)/g

Worked example: v = 20 m/s at 30° → 35.3 m; at 60° → 35.3 m again. The flat shot takes 2.0 s, the lofted one 3.5 s

Open this case: 30°

Complementary angles — two ways to hit the same spot. Two trajectories, one landing point: 30° and 60° are the same range at different heights. Any angle θ and its complement 90° − θ give the same range on level ground, because sin(2θ) = sin(180° − 2θ). — Two trajectories, one landing point: 30° and 60° are the same range at different heights.

Straight up — all height, no range

What you know: θ = 90°, so the horizontal component is zero. Everything the launch gave goes into climbing and coming back.

Range: h = v²/2g, R = 0

Worked example: v = 20 m/s straight up → maximum height v²/2g = 20.4 m, range 0, and it returns to your hand at 20 m/s

Open this case: 90° straight up

Straight up — all height, no range. Vertical launch: the trajectory collapses to a line, and the range vanishes. θ = 90°, so the horizontal component is zero. Everything the launch gave goes into climbing and coming back. — Vertical launch: the trajectory collapses to a line, and the range vanishes.

Launched from a height — the symmetry breaks

What you know: The landing point is below the launch point, so the flight is no longer symmetric and the 45° rule no longer holds.

Range: h > 0 ⇒ θ_opt < 45°

Worked example: v = 22 m/s at 35° from 18 m up → 3.6 s in the air and 65 m downrange; the optimal angle here is 37.2°, not 45°

Open this case: from a cliff

Launched from a height — the symmetry breaks. Extra height buys extra flight time, so a flatter throw wins. The landing point is below the launch point, so the flight is no longer symmetric and the 45° rule no longer holds. — Extra height buys extra flight time, so a flatter throw wins.

Problems solved in full

A ball launched at 20 m/s and 45° from level ground 5 steps
A ball is launched at 20 m/s and 45° from level ground. Find how long it stays up and how far it lands — then find what a 5° aiming error costs, which is the question that decides whether 45° is worth aiming for.
1. Split the launch into two motions that never speak to each other. Gravity acts only downwards, so the horizontal speed is the same at landing as at launch, and 45° is the one angle where the two components start out equal.
  
  $v_{x} = v_{0} cos θ = 20 cos 4 5^{\circ} = 14.142 m/s, v_{y} = v_{0} sin θ = 14.142 m/s$
2. The flight ends when the height comes back to zero, and only the vertical motion knows anything about that. Rise and fall take the same time from level ground, so the whole flight is twice the time to the top.
  
  $y (t_{f}) = 0 : t_{f} = \frac{2 v _{y}}{g} = \frac{2 ( 14.142 )}{9.81} = 2.8832 s$
3. Horizontal distance is one constant speed carried for that flight time. Substituting the components collapses everything into the doubled angle, and since sin 2θ is unchanged when θ is swapped for 90° − θ, launching at 30° or at 60° both land the ball at 35.312 m.
  
  $R = v_{x} t_{f} = 14.142 \times 2.8832 = 40.775 m = \frac{v _{0}^{2} sin 2 θ}{g} = \frac{400}{9.81}$
4. The peak comes from the vertical component alone, brought to rest by gravity. At 45° it lands on exactly a quarter of the range, and that quarter holds at 45° for every launch speed there is.
  
  $y_{m a x} = \frac{v _{y}^{2}}{2 g} = \frac{200}{19.62} = 10.194 m = \frac{R}{4}$
5. Now the aiming error. sin 2θ is stationary at 45°, so missing by 5° is a second-order error and costs 62 cm out of 40.775 m. The same 5° miss at 30° is first-order and moves the range by 8.5%, roughly 3 m.
  
  $\frac{R ( 5 0 ^{\circ} )}{R ( 4 5 ^{\circ} )} = sin 10 0^{\circ} = 0.9848 (- 0.62 m), \frac{R ( 3 5 ^{\circ} )}{R ( 3 0 ^{\circ} )} = \frac{sin 7 0 ^{\circ}}{sin 6 0 ^{\circ}} = 1.085$
Answer

The tool prints 2.8832 s of flight, a range of 40.775 m and a peak of 10.194 m. Those last two are in a ratio of exactly 4, and they stay in that ratio at 45° whatever the speed — a 45° trajectory is always four times as wide as it is tall, so there is one shape and only the size ever changes. Step 5 is the one to keep: the maximum is flat, which is why 45° survives as a rule of thumb that nobody checks carefully. The flatness cuts both ways, though, because a flat peak is a hard peak to find — anyone hunting the best angle by trial gets a wide band of angles that all look equally good.

$R = \frac{v _{0}^{2} sin 2 θ}{g}, \frac{R}{y _{m a x}} = 4 (θ = 4 5^{\circ})$
The angle that carries furthest from 18 m up 7 steps
The panel solves the throw you hand it: 35° from 18 m up carries 64.764 m. It will not tell you whether 35° was worth handing it. Find the angle that carries furthest from this cliff — then ask how much the answer is actually worth.
1. Split the launch in two. Everything after this line is a constant horizontal speed and a vertical free fall that happens to start 18 m above the landing ground.
  
  $v_{x} = v_{0} cos θ = 22 cos 3 5^{\circ} = 18.021 m s^{- 1}, v_{y} = v_{0} sin θ = 22 sin 3 5^{\circ} = 12.619 m s^{- 1}$
2. The flight ends when the height comes back to zero, which is a quadratic in t. Take the positive root; the other one is the time the projectile would have left the ground had it started down there.
  
  $h + v_{y} t - \frac{1}{2} g t^{2} = 0 ⟹ t_{f} = \frac{v _{y} + v _{y}^{2} + 2 g h}{g} = \frac{12.619 + 159.24 + 353.16}{9.81} = 3.5938 s$
3. Range is the horizontal speed multiplied by the whole flight, because nothing acts horizontally at all. Three of the four numbers so far are on the panel above.
  
  $R = v_{x} t_{f} = 18.021 \times 3.5938 = 64.764 m$
4. The peak is where the vertical speed runs out, and the cliff just lifts the whole thing: 18 m of head start plus exactly the rise a ground-level launch would have managed.
  
  $y_{m a x} = h + \frac{v _{y}^{2}}{2 g} = 18 + \frac{159.24}{19.62} = 26.116 m$
5. Now the question the panel does not answer. Write the range as a function of the angle alone, holding v₀ and h fixed. The flight time depends on θ as well now, which is precisely why this is not the tidy sea-level formula.
  
  $R (θ) = \frac{v _{0} cos θ}{g} (v_{0} sin θ + v_{0}^{2} sin^{2} θ + 2 g h)$
6. Differentiate and set to zero. Everything collapses into one condition on sin θ, and the cliff enters only through the dimensionless group 2gh/v₀². Set h = 0 and it returns sin θ = 1/√2, the familiar 45°. Here it returns 37.248°.
  
  $\frac{d R}{d θ} = 0 ⟹ sin θ^{*} = \frac{1}{2 + 2 g h / v _{0}^{2}} = \frac{1}{2.72967} = 0.60526 ⟹ θ^{*} = 37.24 8^{\circ}$
7. Now put the three angles side by side and read the sizes, not just the order.
  
  $R (37.24 8^{\circ}) = 64.887, R (3 5^{\circ}) = 64.764, R (4 5^{\circ}) = 63.355 (m)$
Answer

The best angle from this cliff is 37.248°, and re-aiming is worth 12 cm on a 65 m throw. A maximum is flat, so missing it by 2.2° costs almost nothing. What is expensive is bringing the wrong model — 45° is optimal only from ground level, and using it here throws away 1.532 m, twelve times the cost of the misaim. Raise the cliff and θ* keeps falling; in the limit of a very high one, the furthest throw is very nearly horizontal.

$sin θ^{*} = (2 + \frac{2 g h}{v _{0}^{2}})^{- 1/2} ⟹ θ^{*} = 37.24 8^{\circ}; R^{*} - R (3 5^{\circ}) = 0.123 m, R^{*} - R (4 5^{\circ}) = 1.532 m$

Learning path

Throwing things

Leads to Drag and wind the two-dimensional case where that condition still holds.

References (1)

Insight block 2 stacks two motions. Arriving at that decomposition took Galileo most of a career: S. Drake, "Galileo’s Discovery of the Parabolic Trajectory", in Essays on Galileo and the History and Philosophy of Science, 160–170. University of Toronto Press, 1999.

By the 42Math team · Updated 2026-08-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Physics lessons →

Example problems

45° max range - v=20 m/s, θ=45°, h=0 → range≈40.8 m (maximum for this speed)
30° - v=20 m/s, θ=30° → range≈35.3 m (same as 60° by symmetry)
60° - v=20 m/s, θ=60° → range≈35.3 m (same as 30°)
90° straight up - v=20 m/s, θ=90° → straight up, range=0, max height≈20.4 m
low-angle shot - v=55 m/s, θ=12° → flat trajectory, range≈125 m
from a cliff - v=22 m/s, θ=35°, h=18 m → range≈65 m, flight≈3.6 s