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বাং

Launch It Yourself: How Projectiles Fly

Projectile motion is the curved path a body follows when it is thrown at an angle and moves under gravity alone; the path is a parabola. Change the initial velocity, launch angle and g below and the time of flight, maximum height and range are calculated as it flies.

Drag on the stage to aim: direction sets the angle, length sets the speed

TrajectoryVelocityComponents vₓComponents vᵧEarlier launch
Speed

Controls

20 m/s
45 °
0 m

Gravitational acceleration, g (m/s²)

Readings

Time, t
0.00s
Horizontal distance, x
0.0m
Height, y
0.0m
Resultant velocity, v
20.0m/s
Horizontal velocity, vₓ
14.1m/s
Vertical velocity, vᵧ
14.1m/s
Maximum height, H
10.2m
Horizontal range, R
40.8m
Time of flight, T
2.89s

How to use this simulation

  1. Move the velocity and angle sliders, or drag on the stage to aim directly: the drag direction sets the angle, its length sets the speed.
  2. Keep "earlier launches" on and fire at 30° and then 60°: the two land at the same range.
  3. Switch worlds: with the Moon's weaker g the same throw goes much farther and higher.
  4. The green arrow (vₓ) never changes; the red one (vᵧ) shrinks to zero at the top and flips.
  5. Slow it to 0.25× and pause at any moment to copy the readings into your notebook.

Every throw you have ever made was a projectile

Watch a long pass in football, a basketball free throw or a golf drive. None of them travels in a straight line. The ball rises, seems to hang for a moment, then curves back down. That graceful curve is what this page is about.

Think about tossing your keys to a friend across the room. Throw straight up and they fall back into your own hand. Throw flat and they hit the floor halfway. Without doing any maths you pick an angle, a bit up and a bit forward. That instinct is projectile physics.

A garden hose, a fountain, a frog leaping, a paper ball aimed at the bin: each follows the same rules. Once you know those rules you can predict, before the throw, how high the object will rise, where it will land and how long it will stay in the air.

Starting from zero: two motions that ignore each other

The single best trick in this chapter is to split the motion into two parts: horizontal (sideways) and vertical (up and down). The surprising part is that they do not affect each other at all. Each does its own thing, and together they draw the curve.

Horizontally, once the ball leaves your hand nothing pushes it forward and nothing pulls it back (ignore air resistance for now). Newton's first law says that with no force the velocity stays the same. So the horizontal velocity is constant, and the ball covers equal distances sideways every second.

Vertically, the Earth pulls down the whole time. The upward speed drops by about 9.8 m/s every second until it hits zero at the very top, then the ball starts falling and its downward speed grows by 9.8 m/s every second.

Put them together: steady forward progress plus a rise-and-fall. Distance forward grows in step with time, height changes with the square of time, and a curve built that way is called a parabola. That is why every projectile path (without air resistance) is a parabola.

Try this thought experiment. From the same height, drop one ball and at the same instant flick another sideways. Which lands first? Both land together. Falling is a purely vertical matter, and vertically both start at zero speed with the same g. The sideways speed only carries the second ball farther; it does not change how fast it falls.

Key terms at a glance

Definitions first; most "what is…" exam questions come straight from this table.

TermSymbolPlain meaningUnit
Projectile—An object thrown at an angle that then moves under gravity alone—
Initial velocityv₀The speed at the moment of launchm/s
Launch angleθThe angle the initial velocity makes with the horizontaldegrees (°)
Time of flightTTotal time from launch to landings
Maximum heightHThe highest point reachedm
Horizontal rangeRHorizontal distance from launch point to landing pointm
Trajectory—The curved path the projectile follows; a parabola—
Gravitational accelerationgHow much gravity changes the velocity each second (≈ 9.8)m/s²

Deriving the formulas step by step

If you can derive the formulas yourself you will never mix them up in an exam. Take a body launched from the ground at speed v₀ and angle θ, with no air resistance, and take upward as positive.

Step 1: Split the velocity

The initial velocity is a slanted arrow. Trigonometry splits it into a horizontal and a vertical component:

vₓ = v₀cosθhorizontal component, constant throughout

vᵧ₀ = v₀sinθstarting vertical component

vᵧ = v₀sinθ − gtvertical velocity after time t

Step 2: Time to reach the top

At the top the body stops rising, so its vertical velocity is zero. Setting vᵧ = 0 gives v₀sinθ − gt = 0, so the time to the top is t = v₀sinθ/g.

Going up and coming down happen under the same gravity over the same height, so the descent takes exactly as long. Add the two and you have the time of flight.

t(top) = v₀sinθ / g

T = 2v₀sinθ / gtime of flight

Step 3: Maximum height

Use v² = u² − 2gh vertically. At the top the final velocity is 0 and the initial vertical velocity is v₀sinθ, so 0 = v₀²sin²θ − 2gH.

H = v₀²sin²θ / 2gmaximum height

Step 4: Horizontal range

Horizontally the velocity is constant, so distance = velocity × time. Moving at vₓ for the whole flight gives R = v₀cosθ × 2v₀sinθ/g. Using the identity 2sinθcosθ = sin2θ tidies it up.

R = v₀²sin2θ / ghorizontal range

Step 5: The path equation, and why it is a parabola

From x = v₀cosθ·t we get t = x/(v₀cosθ). Substitute that into y = v₀sinθ·t − ½gt² and time disappears, leaving only x and y.

The result has the form y = ax − bx² with constants a and b. A quadratic in x is a parabola, which proves the shape of the path.

y = x·tanθ − gx² / (2v₀²cos²θ)equation of the trajectory

Step 6: Why 45° goes farthest

With v₀ and g fixed, R depends only on sin2θ. The largest value of sine is 1, reached when 2θ = 90°, that is θ = 45°.

In everyday words: throw too steeply and the ball stays up a long time but hardly moves forward; throw too flat and it moves forward fast but lands almost at once. 45° is the best compromise between the two.

Rmax = v₀² / gat θ = 45°

Step 7: Complementary angles share a range

For θ and 90° − θ, sin2θ and sin(180° − 2θ) are equal because sin(180° − x) = sin x. So 20° and 70°, or 30° and 60°, land at the same spot, although the steeper throw climbs higher and takes longer.

Step 8: Thrown horizontally from a height

Throw a ball straight out (θ = 0°) from a rooftop and its initial vertical velocity is zero. It then falls exactly like a dropped ball, h = ½gt², while moving forward at u the whole time.

t = √(2h / g)time to reach the ground

x = u·√(2h / g)horizontal distance travelled

Experiments to try in the simulation

The animation above is your pocket laboratory. Run these one at a time, and before each one guess the result, then check.

  • Experiment 1: keep the speed at 20 m/s and launch at 15°, 30°, 45°, 60° and 75°. Which angle goes farthest?
  • Experiment 2: with "keep earlier launches" on, compare 30° and 60° side by side. The two paths land on the same point.
  • Experiment 3: keep the angle and double the speed from 10 to 20 m/s. Does the range double or quadruple? The formula has v₀², so it should quadruple.
  • Experiment 4: move from Earth to the Moon. How many times farther does the same throw go? What happens on Jupiter?
  • Experiment 5: raise the launch height h₀. Now 45° is no longer the best angle; from a height a slightly lower angle goes farther.
  • Experiment 6: pause at the top and look for the red arrow (vᵧ). It has vanished, but the green arrow (vₓ) is still there.

Solved problems

Now with numbers. Each solution is worked step by step, and you can check every one by entering the values in the simulation. g = 9.8 m/s² throughout.

Problem 1: A ball thrown at 20 m/s and 45°

Question: a ball is thrown at 20 m/s at 45° above the horizontal. Find the time of flight, the maximum height and the range.

Components first: vₓ = 20 × cos45° = 14.14 m/s and vᵧ₀ = 20 × sin45° = 14.14 m/s.

Time of flight T = 2 × 14.14 / 9.8 = 2.89 s. It reaches the top at half of that, 1.44 s.

Maximum height H = (14.14)² / (2 × 9.8) = 10.20 m, and range R = 20² × sin90° / 9.8 = 40.82 m. Since the angle is 45°, this is the greatest range possible at this speed.

Problem 2: 30° against 60°

Question: the same 20 m/s throw is made once at 30° and once at 60°. Which goes farther?

At 30°: R = 35.35 m, H = 5.10 m, T = 2.04 s. At 60°: R = 35.35 m, H = 15.31 m, T = 3.53 s.

The ranges are identical. The difference is height and time: the 60° ball climbs three times as high and hangs in the air much longer. Which one gives a fielder more time to catch it?

Problem 3: A ball thrown straight out from a roof

Question: a ball is thrown horizontally at 10 m/s from a roof 20 m high. When and where does it land?

Time to fall t = √(2 × 20 / 9.8) = 2.02 s. In that time it moves forward x = 10 × 2.02 = 20.20 m.

At impact the vertical velocity is gt = 19.8 m/s and the horizontal velocity is still 10 m/s, so the resultant speed is √(10² + 19.8²) = 22.2 m/s.

Problem 4: Finding the angle for a given range

Question: you throw at 20 m/s and want to land a ball in a bucket exactly 30 m away. What angle do you use?

From R = v₀²sin2θ/g, sin2θ = Rg/v₀² = 30 × 9.8 / 20² = 0.735. So 2θ = 47.3°, giving θ = 23.7°.

But that is not the whole answer: the complementary angle 66.3° gives exactly the same range. There are two answers: 23.7° (low and fast) or 66.3° (high and slow).

Problem 5: The same throw on the Moon

Question: the ball from Problem 1 is thrown the same way on the Moon (g = 1.62 m/s²). What is the range now?

R = 20² / 1.62 = 246.91 m, H = 61.73 m, T = 17.46 s. That is about 6.0 times the Earth values.

All three formulas have g in the denominator: the weaker the gravity, the farther, higher and longer the flight. Pick "Moon" in the simulation to watch it.

What changes with the angle: a quick table

Values for v₀ = 20 m/s and g = 9.8 m/s² at five angles. Compare the ranges of the pairs 15° and 75°, and 30° and 60°.

  • A bigger angle always means a greater height and a longer flight.
  • The range grows up to 45° and then shrinks, and complementary angles share a range.
Launch angleRange RMaximum height HTime of flight T
15°20.4 m1.4 m1.06 s
30°35.3 m5.1 m2.04 s
45°40.8 m10.2 m2.89 s
60°35.3 m15.3 m3.53 s
75°20.4 m19.0 m3.94 s

Mistakes almost everyone makes

A few errors come up again and again in this chapter. Knowing them saves easy marks in multiple-choice questions.

  • Wrong: "the velocity is zero at the top." Right: only the vertical velocity is zero; the horizontal velocity v₀cosθ is still there.
  • Wrong: "the acceleration is zero at the top." Right: the acceleration is g downward everywhere on the path, including the top.
  • Wrong: "a heavier ball goes less far." Right: without air resistance, mass appears in none of the formulas.
  • Wrong: treating sin2θ as 2sinθ. sin2θ = 2sinθcosθ; sin90° = 1 but 2sin45° ≈ 1.414.
  • Wrong: using T = 2v₀sinθ/g for a launch from a height. That formula only holds when the body lands at its launch height.

Projectiles in real life

Sport is full of projectiles. Corner kicks, basketball shots, the javelin and the shot put: athletes train until they feel the right angle. A shot put leaves from shoulder height, so its best angle is a little below 45°, the effect you saw by raising h₀ in the simulation.

Firefighters choose the angle of a hose by the reach of the water, and fountain designers match the water speed and nozzle angle to shape their arches.

Real air pushes back. Drag slows the body in both directions, so the real path is not a perfect parabola: the descent is steeper, the range shorter and the best angle slips below 45°. Light, bulky objects such as a badminton shuttlecock feel this most. Textbook formulas describe the ideal, drag-free case.

Exam corner

Projectile questions follow a few familiar patterns. Here is what to expect and how to answer.

Definitions

"Define projectile / time of flight / range." Use the precise wording from the key-terms table, and mention "under gravity alone".

Short reasoning

"Why is the path a parabola?" Show the path equation y = ax − bx² and say it is quadratic in x. "Why is the velocity not zero at the top?" Point to the horizontal component.

Calculations

Given v₀ and θ, find T, H or R, exactly like Problem 1. Always write the units.

Higher-order questions

"Will the ball clear a wall 15 m away and 5 m high?" Put x = 15 m into the path equation, find y and compare it with the wall. Or compare two throws at different angles and decide which lands farther.

Quick revision summary

Read just this before an exam and the whole chapter comes back.

  • Projectile: thrown at an angle, moving under gravity alone; the path is a parabola.
  • vₓ = v₀cosθ is constant; vᵧ = v₀sinθ − gt changes; acceleration is always g downward.
  • T = 2v₀sinθ/g, H = v₀²sin²θ/2g, R = v₀²sin2θ/g.
  • The range is greatest at 45° (v₀²/g); complementary angles share a range.
  • Thrown horizontally from height h: t = √(2h/g), x = u·t.
  • Weaker g (the Moon) raises T, H and R; mass appears in no formula.

Frequently asked questions

What is the formula for the range of a projectile?

For a launch and landing at the same height, R = v₀²sin2θ/g, where v₀ is the launch speed, θ the launch angle and g the gravitational acceleration.

Why is the range maximum at 45°?

Range depends on sin2θ, whose largest value is 1 at 2θ = 90°, that is θ = 45°. Then Rmax = v₀²/g. Air resistance lowers the best angle slightly in real life.

What is the time of flight of a projectile?

It is the total time the projectile stays in the air: T = 2v₀sinθ/g when it lands at the height it was launched from.

Is the horizontal velocity of a projectile constant?

Yes. Without air resistance no horizontal force acts, so vₓ = v₀cosθ is constant; only the vertical velocity changes, at the rate g.

What are the velocity and acceleration at the highest point?

At the top the vertical velocity is zero, so the velocity equals the horizontal component v₀cosθ. The acceleration is still g, straight down.

At what angle are the range and maximum height equal?

Setting R = H gives tanθ = 4, so θ ≈ 76°.

Does mass affect projectile motion?

Not without air resistance. Mass appears in none of the formulas for T, H or R, so a heavy and a light ball with the same speed and angle follow the same path.

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The animation made the idea click; now turn it into marks. Syllabus, suggestions, textbooks and admission-test guides are below.

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