(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A water hose is used to fill a large cylindrical storage tank of diameter D and height 2D. The hose shoots water at 45 degrees above the horizontal from the same level as the base of the tank and is at a distance 6D away. For what range of launch speeds will the water enter the tank? ignore air resistance and express answer in terms of D and g

2. Relevant equations

V_{y}=V_{o}sin[tex]\alpha[/tex]-gt

y = (V_{o}sin[tex]\alpha[/tex])t - 1/2 gt^{2}

x = (V_{o}cos[tex]\alpha[/tex])t

v[tex]_{0}[/tex]

3. The attempt at a solution

I see there will be 2 unknowns, time and initial velocity. So I solved for time when the water is at its max height where Vy = 0.

[tex]Vy = Vosin45 - gt[/tex]

[tex]0 = Vo sin45 - gt[/tex]

[tex]gt = Vo sin45[/tex]

[tex]t= (Vo sin45)/g[/tex]

Now, plug t into the height equation to solve for Vo

y = (V_{o}sin[tex]\alpha[/tex])t - 1/2 gt^{2}

2D = (V_{o}sin45)((Vosin45)/g)) - 1/2 g((Vosin45)/g))^{2}

2D = ((V_{o}sin45)^2)/g) - 1/2((Vosin45)^2/g))

2D = 1/2 (((V_{o}sin45)^2)/g)

4Dg = ((V_{o}sin45)^2)

[tex]\sqrt{4Dg}[/tex] = ((V_{o}sin45))

Vo = 2D/sin45

now plug Vo from the previous equation and t into the distance formula for both 6D and 7D to get the range.

for 6D:

6D = ((2Dcos45)/sin45)((Vosin45)/g)

the sin45's cancel...

6D = (2Dcos45/g)Vo

3Dg =Vo cos45

3Dg/Cos45 = Vo

for 7D i did the same and my final result was:

(7/2 Dg)/cos45 = Vo

I don't really know if its right. I have a feeling it won't make it into the container for whatever reason.

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# Homework Help: Projectile motion, water cannon into storage tank

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