Please answer with Solution:

Dear Student,
The solution/s to your 1st, 2nd and 3rd query has/have been provided below:

$$\begin{gathered} \left(1\right) \mathrm{At} \mathrm{maximum} \mathrm{height}, \mathrm{vertical} \mathrm{component} \mathrm{of} \mathrm{velocity}, {\mathrm{v}}_{\mathrm{y}}=0 \\ {\mathrm{u}}_{\mathrm{x}}=\mathrm{ucos\theta } \\ {\mathrm{u}}_{\mathrm{x}}=30\cos 37°=30\times \frac{4}{5}=24 \mathrm{m}/\mathrm{s} \\ {\mathrm{u}}_{\mathrm{y}}=\mathrm{usin\theta } \\ {\mathrm{u}}_{\mathrm{y}}=30\sin 37°=30\times \frac{3}{5}=18 \mathrm{m}/\mathrm{s} \\ \mathrm{a}=-\mathrm{g}=-10 \mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Let} \mathrm{H} \mathrm{be} \mathrm{the} \mathrm{maximum} \mathrm{height} \mathrm{reached} \mathrm{by} \mathrm{rock} \mathrm{measured} \mathrm{from} \mathrm{roof}, \\ \mathrm{s}=\mathrm{H} \\ {{\mathrm{v}}_{\mathrm{y}}}^2={{\mathrm{u}}_{\mathrm{y}}}^2+2\mathrm{as} \\ {0}^2={18}^2-2\times 10\times \mathrm{H} \\ \mathrm{H}=\frac{324}{20}=16.2 \mathrm{m} \\ \left(2\right) \mathrm{to} \mathrm{find} \mathrm{time} \mathrm{of} \mathrm{flight}, \mathrm{t} \\ {\mathrm{s}}_{\mathrm{y}}=-\mathrm{h}=-15 \mathrm{m} \\ {\mathrm{s}}_{\mathrm{y}}={\mathrm{u}}_{\mathrm{y}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ -15=18\mathrm{t}-\frac{1}{2}\times 10{\mathrm{t}}^2 \\ 5{\mathrm{t}}^2-18\mathrm{t}-15=0 \\ \mathrm{t}=\frac{18\pm \sqrt{{\left(-18\right)}^2-4\left(5\right)\left(-15\right)}}{2\times 5} \\ \mathrm{t}=\frac{18\pm \sqrt{324+300}}{10} \\ \mathrm{t}=\frac{18\pm \sqrt{624}}{10} \\ \mathrm{t}=\frac{18\pm 25}{10} \\ \mathrm{t}=\frac{18+25}{10}=4.3 \mathrm{s} \left(\mathrm{ignoring} \mathrm{negative} \mathrm{value}\right) \\ {\mathrm{v}}_{\mathrm{y}}={\mathrm{u}}_{\mathrm{y}}+\mathrm{at} \\ {\mathrm{v}}_{\mathrm{y}}=18-10\times 4.3=-25 \mathrm{m}/\mathrm{s} \\ \mathrm{negative} \mathrm{sign} \mathrm{shows} \mathrm{downward} \mathrm{direction} \\ \left(3\right) \mathrm{Horizontal} \mathrm{range}, \mathrm{R}={\mathrm{u}}_{\mathrm{x}}\mathrm{t} \\ \mathrm{R}=24\times 4.3=103.2 \mathrm{m} \end{gathered}$$
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