Formula: Horizontal throw - parabolic path Height Horizontal distance Initial velocity Initial height
$$y ~=~ -\frac{g}{ 2\,{v_0}^2 } \, x^2 ~+~ y_0$$
$$y ~=~ -\frac{g}{ 2\,{v_0}^2 } \, x^2 ~+~ y_0$$
$$x ~=~ v_0 \, \sqrt{ \frac{2 \, (y_0 ~-~ y) }{g} }$$
$$v_0 ~=~ x \, \sqrt{ \frac{g}{ 2 \, (y_0 ~-~ y) } }$$
$$y_0 ~=~ y ~+~ \frac{ g }{ 2 \, {v_0}^2 } \, x^2$$
$$g ~=~ \frac{ 2 \, (y_0 ~-~ y) }{ x^2 } \, {v_0}^2$$
Height
$$ y $$ Unit $$ \mathrm{m} $$
Current height \( y \) above the ground of a horizontally thrown body. \(y\) is therefore the vertical distance of the body to the ground.
The given formula describes a parabolic trajectory \( y(x) \), which the body follows after being thrown / launched. Using this formula you can calculate the current height \(y\) for each current horizontal position \(x\) of the body.
This formula is a special case of the inclined throw. And the minus sign indicates that the body falls downwards (set as negative direction).
Horizontal distance
$$ x $$ Unit $$ \mathrm{m} $$
Distance of the body from the starting throw position to the current horizontal position \(x\) of the body.
Initial velocity
$$ v_0 $$ Unit $$ \frac{\mathrm m}{\mathrm s} $$
Horizontal velocity (in the body's \(x\) direction) at the time of throwing (or releasing). This is the velocity of the body with which you threw the body horizontally.
It is assumed here that this velocity does not change during the flight, that is, it is constant at all times.
Initial height
$$ y_0 $$ Unit $$ \mathrm{m} $$
This is the height from which the body is released.
Gravitational acceleration
$$ g $$ Unit $$ \frac{\mathrm{m}}{\mathrm{s}^2} $$
A constant acceleration with the value \( g = 9.8 \, \frac{\text m}{\text{s}^2}\). The acceleration states that the thrown body increases its vertical velocity by \( 9.8 \, \frac{\text m}{\text{s}}\) every second. The body is in free fall after release.