Numford
Projectile motion

Projectile motion formulas

T = 2u sin θ / g, R = u² sin 2θ / g and H = u² sin²θ / 2g. At 20 m/s and 30° with g = 10 m/s²: T = 2 s, R = 34.64 m, H = 5 m.

Projectile motion

Drag the launch arrow, or type the speed and angle. Slide the time to see the velocity at that moment. The dots are equal steps of time apart. Air resistance is ignored, as in NCERT.

Time of flight T2 s2u sin θ / g
Range R34.64 mu² sin 2θ / g
Max height H5 mu² sin²θ / 2g
Velocity at 0.5 s18.03 m/svx = 17.32, vy = 5, 16.1° up
Pathy = 0.577x - 0.0167x²y = x tan θ - gx² / (2u² cos²θ)

A ball thrown at 20 m/s and 30 degrees rises 5 m, flies for 2 s and lands 34.64 m away.

g (m/s²)
Show the formulas
Time of flight
T = 2u sin θ / g = 2 × 20 × sin 30° / 10 = 2 s
Range
R = u² sin 2θ / g = 20² × sin 60° / 10 = 34.64 m
Max height
H = u² sin²θ / 2g = 20² × sin²30° / (2 × 10) = 5 m
Longest range
At θ = 45°, R max = u² / g = 40 m. θ and 90° - θ give the same range.
Velocity
vx = u cos θ = 17.32 m/s, vy = u sin θ - gt = 5 m/s at t = 0.5 s, so v = 18.03 m/s
Path
y = x tan θ - gx² / (2u² cos²θ), here y = 0.577x - 0.0167x²
Working

Step by step

  1. Given u = 20 m/s at θ = 30°, with g = 10 m/s².
  2. Split the velocity: ux = u cos θ = 17.32 m/s and uy = u sin θ = 10 m/s.
  3. T = 2u sin θ / g = 2 × 20 × sin 30° / 10 = 2 s
  4. H = u² sin²θ / 2g = 20² × sin²30° / (2 × 10) = 5 m
  5. R = u² sin 2θ / g = 20² × sin 60° / 10 = 34.64 m
  6. Path: y = x tan θ - gx² / (2u² cos²θ), here y = 0.577x - 0.0167x².
What the angle does

Range and height at 20 m/s

  • 15° R = 20 mH = 1.34 m, T = 1.04 s
  • 30° R = 34.64 mH = 5 m, T = 2 s
  • 45° R = 40 mH = 10 m, T = 2.83 s
  • 60° R = 34.64 mH = 15 m, T = 3.46 s
  • 75° R = 20 mH = 18.66 m, T = 3.86 s
Derivation

Why it works

  1. Split the launch velocity: ux = u cos θ along the ground and uy = u sin θ upwards. Gravity only acts downwards, so ux never changes.
  2. Vertically, y = u sin θ t - ½gt². The ball is back on the ground when y = 0, so T = 2u sin θ / g.
  3. Horizontally, x = u cos θ t. In the time T it goes R = u cos θ × 2u sin θ / g = u² sin 2θ / g.
  4. At the top the vertical velocity is zero. From v² = u² - 2gs with v = 0, H = u² sin²θ / 2g.
  5. Taking t = x / (u cos θ) out of the y equation gives the path y = x tan θ - gx² / (2u² cos²θ), a parabola.
Questions

Projectile motion formulas: quick answers

What are the three main formulas of projectile motion?

Time of flight T = 2u sin θ / g, range R = u² sin 2θ / g and maximum height H = u² sin²θ / 2g, for a ball that lands at the height it was thrown from.

What are T, R and H at 20 m/s and 30°?

With g = 10 m/s²: T = 2 s, R = 34.64 m and H = 5 m.

Do these formulas work for a ball thrown from a height?

No. They assume the ball lands at the launch level. From a height h, the time is found from the vertical motion; for a level throw it is t = √(2h/g).

Try your own throw