1i think the water bell will be somewhat like a cylinder......i think
A fountain consists of a small hemispherical rose (sprayer) which lies on the surface of the water in a basin, as illustrated in the figure. The rose has evenly distributed small holes in it, through which water spurts at the same speed in all directions.
What is the shape of the water bell formed by the jets?
Source: 200 PPiP
-
UP 0 DOWN 0 1 13
13 Answers
1
1ok now this was actually my friend's idea...
we see the shape will be either parabola or hyperbola
the water droplet coming out at 45° will have longest range
so the shape of the water bell formed near the end will be decided by the path of this water droplet
and it is that of a parabola
if the shape was like hyperbola then the path near the end must tend towards a tangent(i.e. its asymptote)
but as the end is part of a parabola we can never get asymptote
hence it is a parabola
62Not at all..
infact what you have shown is the trajectory of each "sprinkle" from the water bell..
What you have to show is the final shape of the whole thing..
I mean each of the sprinkles will have a "envelope"
You have to find the equation of that ...
1to finish off this thread
the trajectory of droplet is
y=x tan\theta - \frac{1/2gx^2}{u^2}sec^2\theta
maximum y for a given x will be at
tan\theta=\frac{u^2}{gx}
so we get
y_{env}=x\frac{u^2}{gx}- \frac{1/2gx^2}{u^2}\left(1+\frac{u^4}{g^2x^2} \right)
hence equation of envelop of jet is
y_{env}=\frac{u^2}{g}- \frac{1/2gx^2}{u^2}\left(1+\frac{u^4}{g^2x^2} \right)
66That's correct. To make it clear, the equation is
y=\dfrac{u^2}{2g}-\dfrac{g}{2u^2}\,x^2