Hans Georg Schaathun
August 2016
$$h(t) = 10 - 4{,}9t^2$$
$$h^{-1}(y)$$
$$h(t) = 10 - 4{,}9t^2 = y$$
$$h(t) = 10 - 4{,}9t^2 \quad\quad\quad h^{-1}(y) = \sqrt{\frac{10-y}{4{,}9}}$$
$$h^{-1}(h(t)) = \sqrt{\frac{10-(10-4{,}9t^2)}{4{,}9}} = t$$
$$h(h^{-1}(y)) = 10 - 4{,}9\sqrt{\frac{10-y}{4{,}9}}^2 = y$$