Projectile Motion Playground
Fire a projectile and trace its parabola with range, height and flight time.
About this simulation
Set launch speed, angle and starting height; the path traces out with its velocity components shown at each instant and the apex marked, while range, peak height and time of flight read out live. Try 30° and then 60° to see that complementary angles give the same range on level ground, add a launch height and watch that symmetry break, or place a wall and find the angles that clear it.
How to use it
- Set the Launch speed and Angle sliders and watch the ball trace its path; Range, Peak height and Time of flight appear above the flight.
- Choose Velocity or Both under Show to see the horizontal arrow stay the same length while the vertical arrow shrinks, flips at the top and grows again.
- Set the angle to 30°, note the Range, then try 60° at the same speed: on level ground the two ranges match.
- Raise the Launch height slider, or tap Off a cliff, and see the maximum range no longer occur at 45°.
- Use the Wall distance and Wall height sliders, or the Over the wall and Into the wall presets, to find which angles clear the wall.
What to notice
- The horizontal velocity u cos θ never changes during the flight; only the vertical velocity is changed by gravity.
- At the highest point the vertical velocity is zero, so the speed is at its minimum but not zero.
- On level ground, angles θ and 90° − θ give the same range, and 45° gives the greatest.
- Launching from a height increases range and time of flight and breaks the symmetry of the path.
The formulas
- R = u² sin 2θ / g
- Horizontal range R on level ground for launch speed u at angle θ, with g = 9.81 m/s².
- H = u² sin²θ / 2g
- Maximum height H above the launch point.
- T = 2u sin θ / g
- Time of flight T when the projectile lands at the height it was launched from.
Questions
Why do complementary angles of projection give the same range?
The range on level ground is R = u² sin 2θ / g. Since sin 2θ = sin(180° − 2θ) = sin 2(90° − θ), angles θ and 90° − θ give equal ranges. The steeper shot goes higher and stays up longer but moves forward more slowly. In the simulation, 30° and 60° at 60 m/s both land about 318 m away, with very different peak heights.
What is the velocity of a projectile at its highest point?
At the highest point the vertical component of velocity is zero, but the horizontal component u cos θ is unchanged, so the velocity is u cos θ, directed horizontally. The acceleration there is still g downwards. In the simulation, turn on Velocity under Show and watch the readout v_y pass through zero at the apex while v_x stays fixed.
Why is the horizontal velocity of a projectile constant?
Ignoring air resistance, the only force on a projectile is its weight, which acts vertically. With no horizontal force there is no horizontal acceleration, so the horizontal velocity stays at u cos θ for the whole flight. The vertical motion is independent and changes at g. In the simulation, the horizontal velocity arrow keeps the same length from launch to landing.
Is 45° the angle for maximum range when a projectile is thrown from a height?
No. From a height, the projectile spends extra time falling below the launch level, and a flatter launch makes better use of that time, so the best angle is less than 45°. The higher the launch point compared with u²/g, the lower the best angle. In the simulation, set a Launch height and compare the Range at 45° with slightly lower angles.
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