Câu 1:

Ta có: 

\(\begin{array}{l}
x = {v_0}\cos \alpha t \Rightarrow t = \dfrac{x}{{{v_0}\cos \alpha }}\\
y = {v_0}\sin \alpha t - \dfrac{1}{2}g{t^2}\\
 \Rightarrow y = {v_0}\sin \alpha .\dfrac{x}{{{v_0}\cos \alpha }} - \dfrac{1}{2}g{\left( {\dfrac{x}{{{v_0}\cos \alpha }}} \right)^2}\\
 \Rightarrow y = x\tan \alpha  - \dfrac{g}{{2v_0^2{{\cos }^2}\alpha }}{x^2}\\
 \Rightarrow y = x\tan \alpha  - \dfrac{g}{{2v_0^2}}\left( {1 + {{\tan }^2}\alpha } \right){x^2}
\end{array}\)

Khi vật chạm đất: 

\(\begin{array}{l}
HI.\tan \alpha  - \dfrac{g}{{2v_0^2}}\left( {1 + {{\tan }^2}\alpha } \right)H{I^2} =  - n\\
 \Rightarrow \left( {\dfrac{{g.H{I^2}}}{{2v_0^2}}} \right){\tan ^2}\alpha  - HI.\tan \alpha  + \left( {\dfrac{{g.H{I^2}}}{{2v_0^2}} - n} \right) = 0\\
\Delta  = {\left( { - HI} \right)^2} - 4.\dfrac{{g.H{I^2}}}{{2v_0^2}}.\left( {\dfrac{{g.H{I^2}}}{{2v_0^2}} - n} \right) \ge 0\\
 \Rightarrow 1 - \dfrac{{2g}}{{v_0^2}}\left( {\dfrac{{g.H{I^2}}}{{2v_0^2}} - n} \right) \ge 0\\
 \Rightarrow 1 + \dfrac{{2gn}}{{v_0^2}} \ge \dfrac{{{g^2}.H{I^2}}}{{v_0^4}}\\
 \Rightarrow H{I^2} \le \dfrac{{v_0^4}}{{{g^2}}}\left( {1 + \dfrac{{2gn}}{{v_0^2}}} \right) = \dfrac{{v_0^2}}{{{g^2}}}\left( {v_0^2 + 2gn} \right)\\
 \Rightarrow H{I_{\max }} = \dfrac{{{v_0}}}{g}\sqrt {v_0^2 + 2gn} 
\end{array}\)

Khi đó: 

\(\begin{array}{l}
\tan \alpha {\rm{\;}} = \dfrac{{ - \left( { - HI} \right)}}{{2.\dfrac{{g.H{I^2}}}{{2v_0^2}}}} = \dfrac{{v_0^2}}{{gHI}} = \dfrac{{v_0^2}}{{g.\dfrac{{{v_0}}}{g}\sqrt {v_0^2 + 2gn} }}\\
 \Rightarrow \tan \alpha  = \dfrac{{{v_0}}}{{\sqrt {v_0^2 + 2gn} }}
\end{array}\)

Vậy \(\alpha  = \arctan \dfrac{{{v_0}}}{{\sqrt {v_0^2 + 2gn} }}\)

Câu 2: 

Ta có: 

\(\begin{array}{l}
t = 2.\dfrac{{0 - {v_0}\sin \alpha }}{{ - g}} = \dfrac{{2{v_0}\sin \alpha }}{g}\\
L = {v_0}\cos \alpha .t = {v_0}\cos \alpha .\dfrac{{2{v_0}\sin \alpha }}{g}\\
 \Rightarrow L = \dfrac{{v_0^2\sin 2\alpha }}{g}
\end{array}\)

L max khi \(\sin 2\alpha  = 1 \Rightarrow 2\alpha  = {90^o} \Rightarrow \alpha  = {45^o}\)