Numford
Projectile motion

Time of flight of a projectile

T = 2u sin θ / g. At 30 m/s and 30° with g = 10 m/s², the ball is in the air for 3 s.

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 T3 s2u sin θ / g
Range R77.94 mu² sin 2θ / g
Max height H11.25 mu² sin²θ / 2g
Velocity at 0.75 s27.04 m/svx = 25.98, vy = 7.5, 16.1° up
Pathy = 0.577x - 0.00741x²y = x tan θ - gx² / (2u² cos²θ)

A ball thrown at 30 m/s and 30 degrees rises 11.25 m, flies for 3 s and lands 77.94 m away.

g (m/s²)
Show the formulas
Time of flight
T = 2u sin θ / g = 2 × 30 × sin 30° / 10 = 3 s
Range
R = u² sin 2θ / g = 30² × sin 60° / 10 = 77.94 m
Max height
H = u² sin²θ / 2g = 30² × sin²30° / (2 × 10) = 11.25 m
Longest range
At θ = 45°, R max = u² / g = 90 m. θ and 90° - θ give the same range.
Velocity
vx = u cos θ = 25.98 m/s, vy = u sin θ - gt = 7.5 m/s at t = 0.75 s, so v = 27.04 m/s
Path
y = x tan θ - gx² / (2u² cos²θ), here y = 0.577x - 0.00741x²
Working

Step by step

  1. Given u = 30 m/s at θ = 30°, with g = 10 m/s².
  2. Vertical launch speed uy = u sin θ = 30 × sin 30° = 15 m/s.
  3. Time to the top: uy / g = 15 / 10 = 1.5 s. The way down takes as long.
  4. T = 2u sin θ / g = 2 × 30 × sin 30° / 10 = 3 s
  5. In that time it goes R = u cos θ × T = 25.98 × 3 = 77.94 m.
Derivation

Why it works

  1. Vertically the ball starts at u sin θ and has an acceleration g downwards, so y = u sin θ t - ½gt².
  2. It lands when y = 0 again: t (u sin θ - ½gt) = 0.
  3. t = 0 is the launch. The other answer is T = 2u sin θ / g.
  4. The time up is u sin θ / g and the time down is the same, so the flight is symmetric.
  5. The time depends only on the vertical part of the velocity, not on how fast the ball moves sideways.
Questions

Time of flight: quick answers

What is the formula for the time of flight?

T = 2u sin θ / g, for a ball that lands at the height it was thrown from. It is twice the time taken to reach the top.

How long is a ball at 30 m/s and 30° in the air?

T = 2 × 30 × sin 30° / 10 = 3 s.

Does the time of flight depend on the horizontal speed?

No. It depends only on the vertical part u sin θ and on g. A faster sideways throw at the same vertical speed stays up just as long.

Try your own throw