Answer:
[tex]\theta= (-74.42)^{\circ} C[/tex]
Explanation:
Horizontal speed of water, [tex]v_{xf}=9\ m/s[/tex]
Height, h = -53 (below pool)
We can find firstly the final vertical speed of the water using third equation of kinematics. So
[tex]v^2_{yf}=u^2_{yi}+2(-g)h\\\\v^2_{yf}=2\times -9.8\times -53\\\\v_{yf}=32.23\ m/s[/tex]
Let [tex]\theta[/tex] is the angle where the falling water moving as it enters the pool. So,
[tex]\tan\theta=\dfrac{v_{yf}}{v_{xf}}\\\\=\dfrac{-32.3}{9}\\\\=-74.42^{\circ} C[/tex]
Hence, the angle is (-74.42)°C.
