Numford
Projectile motion

Projectile motion derivation

Split u into u cos θ and u sin θ and treat the two directions apart: T = 2u sin θ / g, R = u² sin 2θ / g, H = u² sin²θ / 2g. At 10 m/s and 30° with g = 10 m/s²: T = 1 s, R = 8.66 m, H = 1.25 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 T1 s2u sin θ / g
Range R8.66 mu² sin 2θ / g
Max height H1.25 mu² sin²θ / 2g
Velocity at 0.25 s9.01 m/svx = 8.66, vy = 2.5, 16.1° up
Pathy = 0.577x - 0.0667x²y = x tan θ - gx² / (2u² cos²θ)

A ball thrown at 10 m/s and 30 degrees rises 1.25 m, flies for 1 s and lands 8.66 m away.

g (m/s²)
Show the formulas
Time of flight
T = 2u sin θ / g = 2 × 10 × sin 30° / 10 = 1 s
Range
R = u² sin 2θ / g = 10² × sin 60° / 10 = 8.66 m
Max height
H = u² sin²θ / 2g = 10² × sin²30° / (2 × 10) = 1.25 m
Longest range
At θ = 45°, R max = u² / g = 10 m. θ and 90° - θ give the same range.
Velocity
vx = u cos θ = 8.66 m/s, vy = u sin θ - gt = 2.5 m/s at t = 0.25 s, so v = 9.01 m/s
Path
y = x tan θ - gx² / (2u² cos²θ), here y = 0.577x - 0.0667x²
Working

Step by step

  1. Given u = 10 m/s at θ = 30°, with g = 10 m/s².
  2. Split the velocity: ux = u cos θ = 8.66 m/s and uy = u sin θ = 5 m/s.
  3. T = 2u sin θ / g = 2 × 10 × sin 30° / 10 = 1 s
  4. H = u² sin²θ / 2g = 10² × sin²30° / (2 × 10) = 1.25 m
  5. R = u² sin 2θ / g = 10² × sin 60° / 10 = 8.66 m
  6. Path: y = x tan θ - gx² / (2u² cos²θ), here y = 0.577x - 0.0667x².
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 derivation: quick answers

How do you derive the time of flight?

Set the vertical displacement u sin θ t - ½gt² to zero. Apart from t = 0, this gives T = 2u sin θ / g.

How do you derive the range?

Multiply the steady horizontal speed u cos θ by T, then use 2 sin θ cos θ = sin 2θ to get R = u² sin 2θ / g.

How do you derive the maximum height?

At the top the vertical velocity is zero. v² = u² - 2gs then gives 0 = u² sin²θ - 2gH, so H = u² sin²θ / 2g.

Try your own throw