Question

[Key Concept: Height above launch level is h = (g/2) t (T − t), so two observations fix T without knowing g or the launch speed.]

$$\begin{matrix} & & h_{1} = u_{y}t_{1} - \frac{1}{2}{gt}_{1}^{2} \\ & {\text{ }h}_{1} + \frac{1}{2}{gt}_{1}^{2} = u_{y}t_{1} & \text{(i)} \\ & {\text{ }h}_{2} = u_{y}t_{2} - \frac{1}{2}{gt}_{2}^{2} & \text{(i)} \\ & {\text{ }h}_{2} + \frac{1}{2}{gt}_{2}^{2} = u_{y}t_{2} & \text{(ii)} \end{matrix}$$

Equation (i)/(ii)

$$\begin{matrix} & \frac{h_{1} + \frac{1}{2}{gt}_{1}^{2}}{{\text{ }h}_{2} + \frac{1}{2}{gt}_{2}^{2}} = \frac{t_{1}}{t_{2}} \\ & {\text{ }h}_{1}t_{2} - h_{2}t_{1} = \frac{g}{2}\left( t_{1}t_{2}^{2} - t_{1}^{2}t_{2} \right) \\ & g = 2\left\lbrack \frac{{\text{ }h}_{1}t_{2} - h_{2}t_{1}}{t_{1}t_{2}^{2} - t_{1}^{2}t_{2}} \right\rbrack \end{matrix}$$

Time of flight

$$\begin{matrix} & T = \frac{2\mu\text{ }g}{\text{ }g} \\ & \text{ }T = \frac{2}{\text{ }g}\left\lbrack \frac{{\text{ }h}_{1} + \frac{1}{2}{gt}_{1}^{2}}{t_{1}} \right\rbrack \\ & T = \frac{2}{t_{1}}\left\lbrack \frac{{\text{ }h}_{1}}{\text{ }g} + \frac{t_{1}^{2}}{2} \right\rbrack \\ & T = \frac{2}{t_{1}}\left\lbrack \frac{{\text{ }h}_{1}\left( t_{1}t_{2}^{2} - t_{1}^{2}t_{2} \right)}{2\left\lbrack {\text{ }h}_{1}t_{2} - h_{2}t_{1} \right\rbrack} + \frac{t_{1}^{2}}{2} \right\rbrack \\ & T = \left\lbrack \frac{h_{1}t_{1}t_{2}\left( t_{2} - t_{1} \right)}{t_{1}\left( {\text{ }h}_{1}t_{2} - h_{2}t_{1} \right)} + t_{1} \right\rbrack \\ & T = \frac{h_{1}t_{2}\left( t_{2} - t_{1} \right) + t_{1}{\text{ }h}_{1}t_{2} - t_{1}^{2}{\text{ }h}_{2}}{{\text{ }h}_{1}t_{2} - h_{2}t_{1}} \\ & \text{ }T = \frac{h_{1}t_{2}^{2} - h_{2}t_{1}^{2}}{{\text{ }h}_{1}t_{2} - h_{2}t_{1}} \end{matrix}$$