Solve Parabolic Motion Problem: Jet of Water from Fire Hose

[PhoenixWright]

I have a problem that means:

The jet of water from a fire hose comes with a vo speed. If the hose nozzle is located at a distance d from the base of a building, demostrate that the nozzle should be tilted at an angle such that $$tan \alpha = \frac{v_0^2}{gd}$$

so that the jet strikes the building as high as possible. At the point where it hits, is the jet is going up or going down?
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I have tried this:

If the jet reaches the building as high as possible:

$$v_y = 0,$$ so $$t = \frac{v_0 sin \alpha}{g}$$

When the jet reaches the building, x = d, so:

$$x = v_0 cos \alpha t \\<br /> <br /> t = \frac {d}{v_0 cos \alpha}$$

I have, therefore:

$$\frac{v_0 sin \alpha}{g} = \frac {d}{v_0 cos \alpha} \\<br /> <br /> \frac{v_0^2}{gd} = \frac{1}{cos \alpha sin \alpha}$$

I should have $$tan \alpha$$ where I have $$\frac{1}{cos \alpha sin \alpha}$$

What is it wrong?

Thank you!

 

[haruspex]

If the jet reaches the building as high as possible:
$v_y = 0$

Certainly that will be the case at the highest point of the trajectory, for a given angle, but that leaves open the possibility that a higher point could be reached at the given distance by using a different angle. In other words, you have assumed the jet will be horizontal at the target.
(Not saying this is wrong, merely that you have not demonstrated it.)

 

[PhoenixWright]

Certainly that will be the case at the highest point of the trajectory, for a given angle, but that leaves open the possibility that a higher point could be reached at the given distance by using a different angle. In other words, you have assumed the jet will be horizontal at the target.
(Not saying this is wrong, merely that you have not demonstrated it.)

Thank you!

Then, I think I should add a condition more (one equation), but I have no idea what it is. If I add $$y = h_0 + v_0sin \alpha t -1/2 gt^2$$, I would have h unknown; and I think I don't have more equations...

 

[haruspex]

Then, I think I should add a condition more (one equation), but I have no idea what it is. If I add $$y = h_0 + v_0sin \alpha t -1/2 gt^2$$, I would have h unknown; and I think I don't have more equations...

You can substitute for t in there using the equation you had from the X direction. It remains to maximise y wrt alpha.

 

[PhoenixWright]

You can substitute for t in there using the equation you had from the X direction. It remains to maximise y wrt alpha.

Thank you! Now I understand it.