Numford
Projectile motion

Equation of trajectory of a projectile

The path is y = x tan θ - gx² / (2u² cos²θ), a parabola. At 10 m/s and 45° with g = 10 m/s² it is y = x - 0.1x².

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 T1.41 s2u sin θ / g
Range R10 mu² sin 2θ / g
Max height H2.5 mu² sin²θ / 2g
Velocity at 0.35 s7.91 m/svx = 7.07, vy = 3.54, 26.57° up
Pathy = x - 0.1x²y = x tan θ - gx² / (2u² cos²θ)

A ball thrown at 10 m/s and 45 degrees rises 2.5 m, flies for 1.41 s and lands 10 m away.

g (m/s²)
Show the formulas
Time of flight
T = 2u sin θ / g = 2 × 10 × sin 45° / 10 = 1.41 s
Range
R = u² sin 2θ / g = 10² × sin 90° / 10 = 10 m
Max height
H = u² sin²θ / 2g = 10² × sin²45° / (2 × 10) = 2.5 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 = 7.07 m/s, vy = u sin θ - gt = 3.54 m/s at t = 0.35 s, so v = 7.91 m/s
Path
y = x tan θ - gx² / (2u² cos²θ), here y = x - 0.1x²
Working

Step by step

  1. Given u = 10 m/s at θ = 45°, with g = 10 m/s².
  2. x = u cos θ t, so t = x / (u cos θ). Put it into y = u sin θ t - ½gt².
  3. y = x tan θ - gx² / (2u² cos²θ), with tan 45° = 1 and g / (2u² cos²θ) = 10 / (2 × 10² × cos²45°) = 0.1.
  4. So the path is y = x - 0.1x².
  5. y = 0 again at x = tan θ / (g / 2u² cos²θ) = 1 / 0.1 = 10 m, the range.
Derivation

Why it works

  1. Horizontally the ball moves at a steady u cos θ, so x = u cos θ t and t = x / (u cos θ).
  2. Vertically y = u sin θ t - ½gt².
  3. Put t into the second equation: y = u sin θ × x / (u cos θ) - ½g x² / (u² cos²θ).
  4. Simplify: y = x tan θ - gx² / (2u² cos²θ).
  5. This is y = bx - cx² with b and c fixed numbers, the equation of a parabola that opens downwards.
Questions

Equation of trajectory: quick answers

What is the equation of trajectory of a projectile?

y = x tan θ - gx² / (2u² cos²θ). It has the form y = bx - cx², which is a parabola.

What is the path at 10 m/s and 45°?

With g = 10 m/s² it is y = x - 0.1x², which is zero again at x = 10 m, the range.

Why is the trajectory a parabola and not a circle?

x grows evenly with time but y has a t² term from gravity. Writing y in terms of x gives an x² term, the mark of a parabola.

Try your own throw