The initial velocity $u=\sqrt{3^2+4^2}=5 \mathrm{m} {\mathrm{s}}^{-1}$ and the angle of projection $\theta ={\tan }^{-1}\frac{4}{3}=53°$.

The equation of trajectory of a projectile is

$$\begin{gathered} y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta } \\ \Rightarrow y=\frac{4}{3}x-\frac{10\times 25}{2\times 25\times 9}{x}^2=\frac{4}{3}x-\frac{5}{9}{x}^2 \\ \Rightarrow y=\frac{1}{9}\left(12x-5x^2\right) \end{gathered}$$

Therefore, $\beta =12$ and $\gamma =-5$.