# Chapter 4 - Problems - Page 150: 42

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.

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