Formula: Oblique Throw Current height Initial height Initial vertical velocity Time Gravitational acceleration
$$y(t) ~=~ y_0 ~+~ v_{\text y0} \, t ~-~ \frac{1}{2}\,g\,t^2$$
$$y(t) ~=~ y_0 ~+~ v_{\text y0} \, t ~-~ \frac{1}{2}\,g\,t^2$$
$$y_0 ~=~ y(t) - v_{\text y0} \, t ~+~ \frac{1}{2}\,g\,t^2$$
$$v_{\text y0} ~=~ \frac{1}{t} \, \left( y(t) ~-~ y_0 ~+~ \frac{1}{2}\,g\,t^2 \right)$$
$$t ~=~ \frac{ v_{\text y0} }{ g } ~\pm~ \sqrt{ \left( \frac{v_{\text y0}}{g} \right)^2 ~+~ \frac{ 2(y_0 - y(t)) }{g} }$$
Current height
$$ y $$ Unit $$ \mathrm{m} $$
Height of a thrown (or launched) body at time \(t\). The height is the vertical distance to the ground.
Initial height
$$ y_0 $$ Unit $$ \mathrm{m} $$
Initial height from which a body is thrown.
Initial vertical velocity
$$ v_{\text y0} $$ Unit $$ \frac{\mathrm m}{\mathrm s} $$
Initial velocity in vertical direction. For example, the body could have been thrown at an angle, which is why it has a velocity \(v_{\text y0}\). The total velocity of the body is \( v_0 = \sqrt{ v_{\text y0}^2 ~+~ v_{\text x0}^2 } \) with \(v_{\text x0}\) as initial velocity in the horizontal direction.
Time
$$ t $$ Unit $$ \mathrm{s} $$
If you insert a time \(t\) into the formula, you get the current height position \(y(t)\) of the body.
Gravitational acceleration
$$ g $$ Unit $$ \frac{\mathrm{m}}{\mathrm{s}^2} $$
Gravitational acceleration is an acceleration of the body due to gravity. On earth it has the value \( g = 9.8 \, \frac{\mathrm m}{\mathrm s^2} \). The minus sign in the formula says that the gravitational acceleration is directed against the \(y\)-axis ("upwards"), which here points against the falling motion.