- #1

CherryWine

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**PROBLEM REGARDING PROJECTILE MOTION**

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I have to find the implicit equation for minimum and maximum initial velocity needed in order for water to end up in the reservoir. I have tried solving it by firstly expressing flight time from the equation for range of projectile (time in terms of D and Vx). I obtain t=6D/(v0*cosx45) - for minimum velocity. Then I substituted this in the equation of y(t) and obtained the trajectory equation. From that equation y(x), I expressed initial velocity and obtained V0=sqrt((D^2*g)/(2(cosx)^2*deltaY-2(cosx)^2*tanx*D)), and from there I obtained Vmin=2,44*sqrt(D*g) and Vmax=2,65*sqrt(D*g) while the correct answers are Vmin=3*sqrt(D*g) and Vmax=3,13*sqrt(D*g). I couldn't figure what was wrong in my method.

Edit: Please note that in the trajectory equation D means the range (6D and 7D), I did not use just one D for calculation. Also in trigonometric functions x is the angle=45 degrees. DeltaY is 2D.

## Homework Equations

D=V0*cosx*t

t=D/(V0*cosx)

deltaY=V0*sinx*t+1/2*gt^2