Numford
Projectile motion

Velocity of a projectile at any time

vx = u cos θ and vy = u sin θ - gt. At 20 m/s and 30° with g = 10 m/s², after 1.5 s the speed is 18.03 m/s, 16.1° below the horizontal.

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 1.5 s18.03 m/svx = 17.32, vy = -5, 16.1° down
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 = 1.5 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. Horizontal: vx = u cos θ = 20 × cos 30° = 17.32 m/s, the same at every moment.
  3. Vertical at t = 1.5 s: vy = u sin θ - gt = 10 - 10 × 1.5 = -5 m/s.
  4. v = √(vx² + vy²) = √(17.32² + (-5)²) = 18.03 m/s
  5. Direction: tan β = vy / vx, so β = 16.1° below the horizontal. The ball is on its way down.
Derivation

Why it works

  1. At any time the horizontal velocity is still vx = u cos θ, because nothing pushes sideways.
  2. The vertical velocity falls steadily: vy = u sin θ - gt. It is zero at the top and negative on the way down.
  3. The speed is v = √(vx² + vy²).
  4. The direction is tan β = vy / vx, measured from the horizontal. The velocity always points along the path.
  5. Speed is least at the top, where it is u cos θ, and it is u again when the ball lands at the launch level.
Questions

Velocity at any time: quick answers

What is the velocity of a projectile at time t?

vx = u cos θ and vy = u sin θ - gt. The speed is √(vx² + vy²) and the direction is tan β = vy / vx.

What is the velocity at 1.5 s for 20 m/s at 30°?

vx = 17.32 m/s, vy = -5 m/s, so v = 18.03 m/s at 16.1° below the horizontal.

Where is the speed smallest?

At the top, where vy = 0 and only u cos θ = 17.32 m/s is left.

Try your own throw