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

## 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.