Answer
The minimum speed at which the archer fish must spit is $3.27~m/s$
The archer fish must spit at an angle of $79.2^{\circ}$ above the horizontal.
Work Step by Step
We can find the vertical component of the initial velocity:
$v_{fy}^2 = v_{0y}^2+2a\Delta y$
$v_{0y}^2 = v_{fy}^2-2a\Delta y$
$v_{0y} = \sqrt{v_{fy}^2-2a\Delta y}$
$v_{0y} = \sqrt{0-(2)(-9.80~m/s^2)(0.525~m)}$
$v_{0y} = 3.21~m/s$
We can find the time to reach a height of $0.525~m$:
$v_{fy} = v_{0y}+at$
$t = \frac{v_{fy}-v_{0y}}{a}$
$t = \frac{0-3.21~m/s}{-9.80~m/s^2}$
$t = 0.328~s$
We can find the horizontal component of the initial velocity if the horizontal displacement must be $0.200~m$ and the travel time is $0.328~s$:
$v_{0x} = \frac{\Delta x}{t} = \frac{0.200~m}{0.328~s} = 0.610~m/s$
We can find the minimum speed at which the archer fish must spit:
$v_0 = \sqrt{v_{0y}^2+v_{0x}^2}$
$v_0 = \sqrt{(3.21~m/s)^2+(0.610~m/s)^2}$
$v_0 = 3.27~m/s$
The minimum speed at which the archer fish must spit is $3.27~m/s$
We can find the angle $\theta$ above the horizontal:
$tan~\theta = \frac{v_{0y}}{v_{0x}}$
$\theta = tan^{-1}(\frac{v_{0y}}{v_{0x}})$
$\theta = tan^{-1}(\frac{3.21~m/s}{0.610~m/s})$
$\theta = 79.2^{\circ}$
The archer fish must spit at an angle of $79.2^{\circ}$ above the horizontal.
