Projectile motion, water cannon into storage tank

[casemeister06]

Homework Statement

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

Homework Equations

V_y=V_osin$$\alpha$$-gt
y = (V_osin$$\alpha$$)t - 1/2 gt²
x = (V_ocos$$\alpha$$)t

v$$_{0}$$

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.

$$Vy = Vosin45 - gt$$
$$0 = Vo sin45 - gt$$
$$gt = Vo sin45$$
$$t= (Vo sin45)/g$$

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

y = (V_osin$$\alpha$$)t - 1/2 gt²
2D = (V_osin45)((Vosin45)/g)) - 1/2 g((Vosin45)/g))²
2D = ((V_osin45)^2)/g) - 1/2((Vosin45)^2/g))
2D = 1/2 (((V_osin45)^2)/g)
4Dg = ((V_osin45)^2)
$$\sqrt{4Dg}$$ = ((V_osin45))
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 = VoI don't really know if its right. I have a feeling it won't make it into the container for whatever reason.

 

[panzerxiii]

I have this same exact problem. How should I go about solving it?

 

[gneill]

The water will follow a parabolic trajectory from the hose, so it will be on the "downside" of the curve when it enters the top of the tank (arching over the front lip). There are two critical trajectories. The first is where the water is hitting the front lip of the vat, and the second is where it is hitting the back lip.

Can you write the equations of motion for projectiles fired at 45 degrees from the specified distance and passing through those points?

You might find it convenient to know that sin(45) = cos(45) = sqrt(2)/2.

 

[zgozvrm]

I solved for time when the water is at its max height where Vy = 0.

$$Vy = Vosin45 - gt$$
$$0 = Vo sin45 - gt$$
$$gt = Vo sin45$$
$$t= (Vo sin45)/g$$

This is not right. For the stream of water to land inside the tank, the water must be on a downward trajectory. Therefore, V_y must be non-zero.

Instead, calculate t based on the horizontal velocity, which we know does not change.

y = (VosinLaTeX Code: \\alpha )t - 1/2 gt2
2D = (Vosin45)((Vosin45)/g)) - 1/2 g((Vosin45)/g))2
2D = ((Vosin45)^2)/g) - 1/2((Vosin45)^2/g))
2D = 1/2 (((Vosin45)^2)/g)
4Dg = ((Vosin45)^2)
LaTeX Code: \\sqrt{4Dg} = ((Vosin45))
Vo = 2D/sin45

In addition, your last step is incorrect. You should have come up with

$$V_o = \frac{2\sqrt{Dg}}{\sin{45^\circ}}$$

which simplifies to [itex]V_o = 2\sqrt{2Dg}[/tex] since<br /> <br /> $$\sin{45^\circ} = \frac{\sqrt{2}}{2}$$[/itex]