Water is flowing in a trapezoidal channel at a rate of q = 20 m3/s. the critical depth y for such a channel must satisfy the equation 0=1-(q^2 b)/(ga_c^3 ) where g is gravity, ac is the cross-sectional area, and b is the width of the channel at the surface. for this case, the width and the cross-sectional area can be related to depth y by b = 3 + y and ac = 3y + 1/2y2. use initial guesses for the location of the root of a=0.5 and b= 2.5. the error on the solution should not exceed 1%.

Make them both to where they have the denominator of 18that makes 3/2 equal 27/18 and 1/9 equal 2/18so now you have  27/18x+5=2/18subtract 5 from the other sideso now you have 27/18x+2/18-5then subtract 2/18 from 27/18 to get 25/18x=-5then divide 25/18 by -5 and your answer will be  -5/18

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.

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