Time of flight

Name: Time of flight
Uploaded: 2026-01-01
Duration: 15 s
Description: T = 2u sinθ/g — the vertical motion alone decides how long the projectile stays up.

T = 2u sinθ/g — the vertical motion alone decides how long the projectile stays up.

Time of flight0 min · Free lecture

In this lesson

T = 2u sinθ/g — the vertical motion alone decides how long the projectile stays up.

The flight time depends only on the vertical component — a level-launch projectile with the same u_y lands at the same instant.

Level ground: total time T = 2u sinθ/g (up time u sinθ/g, down time equal). The time to the top is half the flight time.

On uneven ground, set y = landing height in y = u_y t − ½gt² and solve the quadratic.

T = \frac{2u\sin\theta}{g}

Time of flight on level ground

Worked example

For a fixed launch speed, the horizontal range of a projectile is maximum when the launch angle is:

R = v0² sin 2θ / g
R is maximum when sin 2θ = 1, i.e. 2θ = 90°, so θ = 45°

Answer: 45°

Transcript (0 min)

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Time of flight — FemtoLearn.
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T = 2u sinθ/g — the vertical motion alone decides how long the projectile stays up.

Printable notes (free)

Practice

Free · 4 questions with full solutions

Q1 · Numerical · difficulty 2/5

A projectile is launched with speed 25 m/s at an angle 15°. Taking g = 9.8 m/s², find its total time of flight.

Q2 · Numerical · difficulty 2/5

A projectile is launched with speed 15 m/s at an angle 70° above the horizontal. Taking g = 10 m/s², find its horizontal range.

R = \frac{v_0^2 \sin 2\theta}{g}

Q3 · Numerical · difficulty 2/5

A projectile is launched with speed 15 m/s at an angle 25°. Taking g = 10 m/s², find its maximum height above the launch point.

H = \frac{v_0^2 \sin^2\theta}{2g}

Q4 · MCQ · difficulty 1/5

For a fixed launch speed, the horizontal range of a projectile is maximum when the launch angle is:

30°
45°
60°
90°