Formula Oblique Throw - Trajectory Height Angle Initial height Horizontal distance Initial velocity
$$y(x) ~=~ y_0 ~+~ \tan(\varphi_0) \, x ~-~ \frac{g}{2{v_0}^2 \, \cos(\varphi_0)^2 } \, x^2$$ $$y(x) ~=~ y_0 ~+~ \tan(\varphi_0) \, x ~-~ \frac{g}{2{v_0}^2 \, \cos(\varphi_0)^2 } \, x^2$$ $$y_0 ~=~ y ~-~ \tan(\varphi_0) \, x ~+~ \frac{g}{2{v_0}^2 \, \cos(\varphi_0)^2 } \, x^2$$ $$x ~=~ \frac{ {v_0}^2 \, \cos(\varphi_0)^2 \, \tan(\varphi_0) }{g} ~\pm~ \sqrt{ \left( \frac{ {v_0}^2 \, \cos(\varphi_0)^2 \, \tan(\varphi_0) }{g} \right)^2 ~-~ \frac{2{v_0}^2 \, \cos(\varphi_0) \, (y-y_0)}{g} }$$ $$v_0 ~=~ \sqrt{ \frac{ g }{ 2\,(y_0 - y + \tan(\varphi_0)\,x) } } \, \frac{x}{ \cos(\varphi_0) }$$
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.