- Launch speed
- 20m/s
- Launch angle
- 45°
- Gravity
- 9.81m/s²
2.883s
Open with these values2.883s
Result: 2.883 sOnly the vertical part of the launch speed keeps a projectile up, so the flight lasts 2 × v × sin θ ÷ g. At 20 m/s and 45° on Earth that is 2.883 s. Gravity is an input, not a fixed number: set it to 1.62 for the Moon or 3.71 for Mars and the same throw hangs far longer.
2.883s
Open with these values5.297s
Open with these values21.824s
Open with these valuest = 2 × v × sin θ ÷ g
| Speed, angle, gravity | Where | Time of flight (s) |
|---|---|---|
| 10, 90, 9.81 | Earth, straight up | 2.039 |
| 20, 45, 9.81 | Earth | 2.883 |
| 50, 30, 9.81 | Earth, fast and flat | 5.097 |
| 30, 60, 9.81 | Earth, steep | 5.297 |
| 25, 45, 1.62 | Moon | 21.824 |
Multiply twice the launch speed by the sine of the angle and divide by gravity: time = 2 × v × sin θ / g. For a 20 m/s launch at 45° on Earth (g = 9.81), that is 2 × 20 × sin 45° / 9.81 ≈ 2.883 s. The time depends only on the vertical part of the velocity, v·sinθ.
Yes — just change the gravity value. Earth is 9.81 m/s², the Moon about 1.62, Mars about 3.71, and Jupiter about 24.79. Weaker gravity means a longer time of flight for the same launch, which is why a throw hangs so long on the Moon.
No. Mass cancels out of the equations of motion, so a pebble and a cannonball launched at the same speed and angle stay up equally long. Only air resistance, which this model ignores, makes the light one behave differently.
Exactly half. The rise takes v × sin θ / g and the fall takes the same again, because launch and landing are at the same height. The peak of the arc therefore sits at the midpoint of the flight time.
No. It uses the ideal, drag-free equations, assuming the projectile launches and lands at the same height with gravity as the only force. Real projectiles stay up slightly less long, so treat these results as the no-drag upper bound.
Information, not professional advice.
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