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Formula Oblique Throw - Trajectory Height    Angle    Initial height    Horizontal distance    Initial velocity   

Formula
Formula: Oblique Throw - Trajectory
Oblique throw - trajectory with initial velocity, angle and height
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Height

Unit
Current height \(y\) of the body dropped at an angle \(\varphi_0\) from the initial height \(y_0\) at the velocity \(v_0\) and. The body is currently located at the horizontal position \(x\).

Angle

Unit
Launch angle between the direction of the initial velocity \(v_0\) and the horizontal (i.e. the \(x\) axis). The launch angle determines how far the body flies. If you manage to throw the body at an angle of \( \varphi_0 = 45 ^{\circ} \), then you will achieve the greatest horizontal distance - for a given initial velocity \( v_0 \).

At an angle of \( \varphi_0 = 0 ^{\circ} \) you throw the body horizontally. The formula then simplifies to a horizontal throw. The index 0 should indicate that it is the angle at the start time \( t = 0 \) of the throw.

Initial height

Unit
Height of the body above the ground at the time you released / launched the body. For example, if you dropped the body from your shoulder, then \(y_0\) is the height from the ground to your shoulder.

Horizontal distance

Unit
Distance of the body from the launch position to the current horizontal position of the body.

Initial velocity

Unit
Constant velocity of the body with which you launched the body at an angle \(\varphi_0\). The index 0 is to indicate that it is the velocity at the time \( t = 0 \) of the launch.

Gravitational acceleration

Unit
A constant acceleration with the value \( g = 9.8 \, \frac{\mathrm m}{\mathrm{s}^2}\). This states that the thrown body increases its vertical velocity by \( 9.8 \, \frac{\mathrm m}{\mathrm{s}}\) every second. The body is in free fall after release.