Projectile Motion: Range, Launch Angle and Time of Flight
A 45-minute guided lesson: students launch projectiles in the simulator, discover that horizontal and vertical motion are independent, find the angle that gives maximum range, and test their predictions on the Moon.
Learning objectives
- Describe projectile motion as two independent motions: constant velocity horizontally and constant acceleration vertically.
- Use the simulator to measure range, maximum height and time of flight and relate them to launch speed and angle.
- Explain why complementary launch angles (e.g. 30° and 60°) give the same range and why 45° gives the maximum range on level ground.
- Predict how range changes when gravity changes, and check the prediction quantitatively.
Students should already know
- Kinematics in one dimension: v = v_0 + a t and x = x_0 + v_0 t + tfrac12 a t^2.
- Resolving a vector into components with sine and cosine.
Standards
- NGSS HS-PS2-1
- AP Physics 1 · Unit 1 (Kinematics)
- IB Physics · Topic A.1
- Common Core HSF-TF.B
1.Make a prediction
~5 minA ball is thrown from the ground with the same speed every time, but at different angles. Before touching anything, write down two predictions:
2.Two motions in one
~10 minThe simulator below is set to a launch speed of 25 m/s at 45° on Earth. The blue arrow is the velocity; the green and orange dashed arrows are its horizontal (v_x) and vertical (v_y) components.
Your task
- Press Launch and watch the two component arrows. Use Pause (or the P key) near the top of the arc.
- Record Range, Max Height and Time of Flight from the readouts on the right.
- Look at the v_x(t) and v_y(t) graphs under the canvas. Which one is a flat line? Which one crosses zero, and where is the ball at that moment?
You should see that the green horizontal component never changes while the orange vertical component shrinks to zero at the top and then grows again pointing downward. Gravity acts only vertically, so only v_y is affected. This is the single most important idea of the lesson: the horizontal and vertical motions are independent, and they share only one thing — time.
Components of the launch velocity
v_0x = v_0 cosθ, quad v_0y = v_0 sinθ
Horizontal motion (no acceleration)
x(t) = v_0 cosθ · t
Vertical motion (acceleration −g)
y(t) = v_0 sinθ · t - tfrac12 g t^2
3.Which angle goes farthest?
~12 minNow test your first prediction. This instance is set to 20 m/s with *ghost trails* on, so every launch stays on the screen and you can compare paths directly.
Your task
- Launch at 30° and write down the range and the time of flight.
- Set the angle to 60° (drag the slider or type the value) and launch again. Compare the range and the time of flight with the 30° launch.
- Repeat for 45°, then 15° and 75°. Fill in a table: angle → range → max height → time of flight.
Two things should stand out. First, 30° and 60° land in the same place (35.3 m), as do 15° and 75° — even though the 60° ball flies much higher and much longer. Second, 45° gives the largest range (40.8 m at 20 m/s). Both facts follow from one formula.
Time of flight (level ground)
T = (2 v_0 sinθ)/(g)
Range
R = v_0 cosθ · T = (v_0^2 sin 2θ)/(g)
Maximum height
H = (v_0^2 sin^2θ)/(2g)
The range depends on the angle only through sin 2θ. Since sin(2 · 30°) = sin 60° and sin(2 · 60°) = sin 120° are equal, the ranges are equal: angles that add up to 90° are called complementary and always share a range. The sine function peaks at 90°, so 2θ = 90°, i.e. \( \theta = 45° \), gives the maximum.
The time of flight and the maximum height, on the other hand, depend on sinθ alone, and that keeps growing up to 90°. So the high 60° shot takes longer and climbs higher than the 30° shot — it simply spends that extra time going up and down rather than forward.
4.Same throw on the Moon
~8 minOn Earth a 25 m/s, 45° throw lands 63.7 m away. Lunar gravity is g = 1.62\ textm/s^2, about one sixth of Earth’s. Before launching, use the range formula to predict the lunar range and the time of flight.
Your task
- Write down your predicted range and time of flight for the Moon.
- Press Launch and compare. Divide the measured lunar range by the Earth range (63.7 m). What number do you get, and why?
- Switch the preset back to Earth and then to Jupiter (24.79). Does the horizontal speed change between planets? Does the time of flight?
Both R and T are inversely proportional to g: the Moon gives 385.8 m and 21.8 s, exactly 9.81 / 1.62 approx 6.06 times the Earth values. The horizontal speed is the same everywhere — the ball flies six times farther only because it stays in the air six times longer.
5.Putting the equations to work
~5 minA kicked ball
A football is kicked from the ground at 18 m/s at an angle of 35° above the horizontal. Ignoring air resistance (g = 9.81 m/s²), find (a) the time of flight, (b) the range, (c) the maximum height.
- Components: v_0x = 18 cos 35° = 14.7\ textm/s, v_0y = 18 sin 35° = 10.3\ textm/s.
- (a) The ball lands when y = 0: T = 2 v_0y / g = 2 · 10.3 / 9.81 = 2.10\ texts.
- (b) Horizontal motion is uniform: R = v_0x T = 14.7 · 2.10 = 31.0\ textm. Check with R = v_0^2 sin 70° / g = 324 · 0.940 / 9.81 = 31.0\ textm.
- (c) At the top v_y = 0: H = v_0y^2 / (2g) = 10.3^2 / 19.62 = 5.4\ textm.
T = 2.10 s, R = 31.0 m, H = 5.4 m. Verify in the simulator: set 18 m/s and 35° — the readouts should match to one decimal.
6.Check your understanding
Answer each question, then press Check. Numeric answers from the simulator are accepted within the stated tolerance.
1.While a projectile is in flight (no air resistance), its horizontal velocity component…
2.Two balls are launched with the same speed, one at 20° and one at 70°. Which statement is true?
3.In Explore 2 you launched at 20 m/s. Using R = v₀² sin 2θ / g, what range should the simulator report at 45°? (g = 9.81 m/s²)
4.A projectile is fired at 30 m/s at 50°. How long is it in the air? (g = 9.81 m/s²)
5.Which of these change when the same throw is repeated on the Moon instead of Earth? (Select all that apply.)
6.At the highest point of its path, a projectile’s acceleration is…
7.Homework
- Find two launch angles that give a range of exactly 30 m at 20 m/s. Verify both in the simulator and explain why there are two answers. Open simulator with these settings ↗
- Turn on quadratic air drag (mass 0.4 kg, k = 0.02). Is 45° still the best angle? Try 35°, 40° and 45° and write one sentence about what drag does to the optimum. Open simulator with these settings ↗
- On Jupiter (g = 24.79 m/s²), how fast must you throw at 45° to reach the same 63.7 m that 25 m/s gives on Earth? Solve R = v₀² / g for v₀, then test your answer against the target marker. Open simulator with these settings ↗
Notes for teachers
- Pacing: Warm-up 5 min, Explore 1 10 min, Explore 2 12 min, Explore 3 8 min, worked example 5 min, quiz 5 min. If time is short, drop Explore 3 and assign it as homework — it only needs the formula.
- Each embedded simulator starts with the settings the step needs; students can still change any slider. The “Open full simulator” link in each frame gives a shareable page with the same settings for projecting or homework.
- Misconceptions this lesson targets: (1) “the ball speeds up horizontally on the way down” — refuted by the v_x(t) graph; (2) “acceleration is zero at the top” — addressed in quiz question 6; (3) “higher angle always means farther” — refuted directly by the 30°/60° comparison.
- Differentiation: stronger students can derive R = v₀² sin 2θ / g from the two kinematic equations before Explore 2 instead of being given it. Students who struggle with trigonometry can work entirely from the readouts and the table in Explore 2.
- Assessment: the numeric quiz questions accept ±3 %, so answers read from the simulator (which integrates numerically) are accepted alongside formula answers. Question 3 can be answered only by a student who actually ran the 45° case or did the calculation.
- Air drag is deliberately off until the homework. With drag on, the complementary-angle result and the 45° optimum both break down, which is a good extension discussion but confuses the first pass.