This is from Khan Academy.
PROBLEM: Alain throws a stone off a bridge into a river below. The stone's height (in meters above the water), X seconds after Alain threw it, is modeled by:
$$h(x)=−5x^2+10x+15$$
What is the height of the stone at the time it is thrown?
SOLUTION: The height of the stone at the time it is thrown is given by h(0). Hence, answer is 15 meters.
I can not understand how h(0) is height of the stone at the time it is thrown.
When I throw a stone off a bridge down into the river, I will throw it straight or throw it downwards. It will only form half of the parabola. Then in this case, the bridge will be the vertex and hence will be the maximum point and hence y co-ordinate of vertex will be the height of bridge and hence height of the stone at the time it is thrown. Formula for finding vertex is:
x = -b/2a = 1, hence h(1) = 10 meters.
But my answer is wrong.
Best Answer
Regardless of whether the stone is initially thrown upwards, downwards, or horizontally, its initial height would still be $h(0)$, which represents the height at which it is released.
As for the physics, the height $h(t)$ of an object at time $t$ that has initial vertical velocity $v_0$ and initial height $h_0$ is $$h(t) = \frac{1}{2}gt^2 + v_0t + h_0$$ where $g = -9.8~\text{m}/\text{s}^2$ is the acceleration due to gravity in the Earth's gravitational field. The author opted to round $-9.8$ to the nearest integer. From the formula, we see that the initial upward velocity is $v_0 = 10~\text{m}/\text{s}$ and the initial height is $h_0 = 15~\text{m}$.
If we complete the square, we obtain \begin{align*} h(x) & = -5x^2 + 10x + 15\\ & = -5(x^2 - 2x) + 15\\ & = -5(x^2 - 2x + 1) + 5 + 15\\ & = -5(x - 1)^2 + 20 \end{align*} which is the equation of a downward facing parabola with vertex $(1, 20)$, so it is true that the stone reaches its maximum height one second after it is released. However, the maximum height is $h(1) = 20~\text{m}$.