Enter the projectile angle, initial height and velocity in the time of flight calculator and the tool will calculate the time of flight.
The time of flight is the time taken to travel at a selected trajectory through the air. launch angle and preliminary velocity decide the time of flight of an item.
The system in the time of flight preliminary pace and angle of the projection of flight: The time of flight equation is given via:
$$ t = \dfrac{V_o sin(α) + \sqrt{(V_o sin(α))^2 + 2gh}}{g} $$
in which:
V_o = Intail speed
α = attitude of Projection
g = Gravitational force
let's think the attitude of projection of an item is 45 degrees, the preliminary height and pace of the object are 5 m and 30 m/sec respectively. The gravitational pull is nine.80665 m/sec^2, then find how lengthy an item is inside the air.
Given:
V_o = 30 m/sec
α = 45 degrees
h = 5m
g = 9.80665 m/sec^2
Solution:
\( t = \dfrac{V_o sin(α) + \sqrt{(V_o sin(α))^2 + 2gh}}{g}\)
Enter the values in the time flight equation, for simplification use the time of flight calculator.
\(t = \dfrac{30 \times sin(45) + \sqrt{(30 \times sin(45))^2 + 2 \times9.807\times5}}{9.807}\)
\( t = \dfrac{30 \times 0.7071 + \sqrt{(30 \times 0.7071)^2 + 2 \times9.807\times5}}{9.807}\)
\(t = \dfrac{21.21 + \sqrt{548.0665}}{9.807}\)
\( t = \dfrac{44.62}{9.807}\)
\( \text{t = 4.55 sec}\)
The time of flight of the item is four.fifty five sec whilst checked through the projectile movement time calculator
| Property | Description | Formula | Example |
|---|---|---|---|
| Time of Flight (T) | Total time a projectile remains in the air. | T = (2 × v₀ × sinθ) / g | If v₀ = 20 m/s, θ = 45°, g = 9.81 m/s², then T ≈ 2.87 s. |
| Initial Velocity (v₀) | Speed at which the projectile is launched. | v₀ = (T × g) / (2 × sinθ) | If T = 3 s, θ = 30°, g = 9.81 m/s², then v₀ ≈ 17.3 m/s. |
| Angle of Launch (θ) | Angle at which the projectile is launched. | θ = sin⁻¹((T × g) / (2 × v₀)) | If T = 4 s, v₀ = 25 m/s, g = 9.81 m/s², then θ ≈ 47.2°. |
| Acceleration due to Gravity (g) | Downward force acting on the projectile. | g = (2 × v₀ × sinθ) / T | If v₀ = 30 m/s, θ = 60°, T = 5 s, then g ≈ 9.81 m/s². |
| Maximum Height (H) | Highest point reached by the projectile. | H = (v₀² × sin²θ) / (2 × g) | If v₀ = 25 m/s, θ = 45°, g = 9.81 m/s², then H ≈ 15.9 m. |
| Range (R) | Horizontal distance traveled by the projectile. | R = (v₀² × sin(2θ)) / g | If v₀ = 20 m/s, θ = 30°, g = 9.81 m/s², then R ≈ 35.3 m. |
| Horizontal Velocity (vₓ) | Constant velocity in the x-direction. | vₓ = v₀ × cosθ | If v₀ = 25 m/s, θ = 45°, then vₓ ≈ 17.7 m/s. |
| Vertical Velocity (vᵧ) | Velocity in the y-direction at any point. | vᵧ = v₀ × sinθ - g × t | If v₀ = 30 m/s, θ = 60°, g = 9.81 m/s², t = 2 s, then vᵧ ≈ 11.8 m/s. |
| Final Velocity (v) | Velocity of the projectile before impact. | v = √(vₓ² + vᵧ²) | If vₓ = 17.3 m/s, vᵧ = 12.1 m/s, then v ≈ 21.2 m/s. |
| Impact Angle (θₑ) | Angle at which the projectile hits the ground. | θₑ = tan⁻¹(vᵧ / vₓ) | If vₓ = 15 m/s, vᵧ = 10 m/s, then θₑ ≈ 33.7°. |
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