91Ó°ÊÓ

It is given that,

$\mathrm{Ï•}=36.0^o$

The problem deals with the projectile motion of an object. The projectile motion of an object is the motion of an object thrown into the air, subject to the acceleration due to gravity Though the drop is launched at a different angle, its launching point is the same as that of the archerfish. So, they have to travel the same distance. Using this, the angle can be found.

Formulae:

The range of the projectile is given by,

$R=\frac{\left(v_0^2\sin n2{\mathrm{θ}}_0\right)}{g}$

Maximum height is given by,

$H_{max}=\frac{\left(v_0^{2}si{n}^2{\mathrm{θ}}_0\right)}{2g}$

From the diagram,

$$\begin{gathered} Y=H=\frac{\left(v_0^2{\sin }^2{\mathrm{θ}}_0\right)}{2g} \\ x=\frac{R}{2}=\frac{\left(v_0^2\sin 2{\mathrm{θ}}_0\right)}{2×g} \\ \tan \mathrm{ϕ}=\frac{Y}{H} \\ \tan \mathrm{ϕ}=\frac{\frac{\left(v_0^2{\sin }^2{\mathrm{θ}}_0\right)}{2g}}{\frac{\left(v_0^2\sin 2{\mathrm{θ}}_0\right)}{2×g}} \\ \tan \mathrm{ϕ}=\frac{1}{2}×\frac{\sin {\mathrm{θ}}_0}{\cos {\mathrm{θ}}_0} \\ \tan \mathrm{ϕ}=\frac{1}{2}×\tan {\mathrm{θ}}_0 \\ \tan \mathrm{ϕ}=2×\tan \mathrm{ϕ} \\ {\mathrm{θ}}_0={\tan }^{-1}(2×\tan \mathrm{ϕ}) \\ {\mathrm{θ}}_0={\tan }^{-1}(2×\tan 36.0) \\ {\mathrm{θ}}_0={\tan }^{-1}(1.453) \end{gathered}$$

Thus,

$$\begin{gathered} {\mathrm{θ}}_0=55.46 \\ ≈55.5° \end{gathered}$$