I don't have a proof, but I do have a derivation based on the physics.
Let's assume a rock is tossed with initial speed U at angle theta above the ground and launched from height h above the ground. We want the vertical height y(t) above the ground at any time t since being tossed.
So first of all it's at y(t = 0) = h before we toss it. And that's our first term in the equation. Then we fling it with Ux = U cos(theta) X speed and Uy = U sin(theta) Y speed.
And that means we have y(t) = h + Uy t as the height after t time of vertical flight at speed Uy. But, a big BUT, there is the gravity of the situation. So that rock does not continue unabated at Uy, it starts to slow down due to gravity.
And that's our next term, the gravity term 1/2 gt^2 which if you will recall is the distance traveled while in a constant acceleration over time t. But this gravity distance is actually a loss of distance since the rock is decelerating at g the gravity rate.
And that means we need to subtract 1/2 gt^2 from y(t) = h + Uy t to get y(t) = h + Uy t - 1/2 g t^2, which is your h = -4.9t^2 + v0t + h0, note that 1/2 g = 4.9. QED.