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

## Height

`$$ y(x) $$`Unit

`$$ \mathrm{m} $$`

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

`$$ \varphi_0 $$`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

`$$ y_0 $$`Unit

`$$ \mathrm{m} $$`

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

`$$ x $$`Unit

`$$ \mathrm{m} $$`

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

## Initial velocity

`$$ v_0 $$`Unit

`$$ \frac{\mathrm m}{\mathrm s} $$`

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

`$$ g $$`Unit

`$$ \frac{\mathrm{m}}{\mathrm{s}^2} $$`

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.