Finding initial velocity of a water fountain?

Question

Assuming the height from the bottom to the water fountain spout(the point that actually releases water) was "H", distance the water travels horizontally is "F", highest point the water reaches is "P", and time it takes from the spout to the bottom of the pan is "T".

I know that I have to use these equations and use either Sin, Cos, or Tan to find the apex.
V=d/t
A=deltaV/T
v=initialV +at
v^2=initialv^2 + 2ad
d=initald +initalv(t) + 1/2 at^2

Some steps on how I would go on solving the problem would be helpful.
Thanks in advance!

Where does the 19.6 come from?
:O

Answer 1

You'll want to break this problem into three parts. (1) Find the initial velocity in the vertical (y) direction. (2)Find the velocity in the horizontal (x) direction. (3) Find the total velocity using Pythagoras' Theorem.

First off, I'm not going to consider the force of air resistance for simplicity. It is possible to take into account air resistance, but thats probably a huge, messy, waste of time here.....

In the first part, use the energy method, i.e. equate the initial kinetic energy, to the potential energy at the stream's highest point:

(1/2)m v0[y]² = mgP.

Then, eliminating "m", and solving:

v0[y] = sqrt(2gP).

The second part is a little trickier. There are no no forces or accelerations in the horizontal, so the velocity stays constant. Therefore You can use the distance the stream travels from the spout to find the horiz. velocity, but you need to know the total "flight time" first.

Since the acceleration of gravity is constant, you can write the y position as a parametric function of time.

y = y0 + v0[y] t - (1/2)g t²,

or, substituting the known values for "y0", and v0[y]:

0 = H + sqrt(2gP) t - (g/2) t²;

(since, at the end of it's travel, y = 0.) The total "flight time" corresponds to the right-hand root, given by the quadratic formula. (Mind you, we already know that t = T, but I'll still work through this part anyway......)

T1, T2 = {-b ± sqrt(b² - 4ac)}/2a,

or, by substitution

t = { sqrt( 2gP + 2Hg) - sqrt(2gP) } / -g.

Since the position in the horizontal is simply

x = v0[x] t,

then

v0[x] = x / t = F / t

or

v0[x] = -Fg / { sqrt( 2gP + 2Hg) - sqrt(2gP) }.

Note: factoring out sqrt(g) simplifies this a bit. Since g = sqrt(g)²;

v0[x] = -F sqrt(g) / { sqrt( 2[P+H]) - sqrt(2P) }

From there, finding the total initial velocity, and the angle should be easy, using

||v||² = v[x]² + v[y]² ,

and

tan θ = v[y] / v[x].

Hope that isn't too confusing,
~W.O.M.B.A.T.

Answer 2

you don't need trig equations, just basic algebra(it forms a parabola)

rad(19.6(P-H)) = initial velocity

This works because, everything falls at the same rate when in free fall and the initial velocity(for the spout) is the same as when the falling water passes the same height as which it left. It's more simple than you think, you don't need any trig equations or any different equations for that matter than what you already have.

eek! sorry about that. I got 19.6 from:

V^2 = Vo^2 + 2AD

A = 9.8m/s^2

So,

V^2 = 0^2 + 2(9.8)D

V^2 = 19.6D

V = rad(19.6D)

D = P - H

V = rad(19.6(P - H))

Sorry I didn't explain, but I promise it's right. I think I got the thumbs down cause I didn't explain where I got it. I tend to do math in my head and just put answers. Bad habit. I appologize. Good Luck! :)

The guys below got way too far into it. You don't need trig equations because everything falls at the same rate. It doesn't matter whether it's a speeding bullet or a spaghetti noodle, they'll hit the ground at the same time. So I just took found what the speed was when it passes the spout from the highest point at free-fall since the velocity of the water passing the spout is the exact same as the initial velocity.

Answer 3

The first thing is find the angle which is

arctan P/F

then find the hieght then use that to find distance it travels: i.e. the hypotuse of the angle formed by F and P. You can actually do this three ways. one is using sin function of the angle theta. one is cos of the angle theta or using using the sum of the squares equal the sqaure of the hypotuse.

hypotuse^2 = F^2 + P^2 but not recommended unless necessary: Too long winded:

Cos theta = cos ( arctanF/P)

distance the water travels to it's apex is D

D = Apex - H or distance is = to P / ( cos(artan F/P) )

V = d/t
V= velocitty
d= distance
t = time

V = ( P / (cos(arctanF/P))) / t .: there is your base formula.
That should get you started. I used a lot of paranthesis not necessary and used arctan instead of tan^-1 because no way to show it correctly
Now all you have to do is use your Trig identities to simplify the equation.
Or defferentiate it to simplify it. further.

You can also use sin function instead of cos function as well. just in case you need to use the opposite versus the adjactent side.

Answer 4

yes you can use cos and sin identities using the formulas in projectile motion
Sx=VocosthetaxT
Sy=VosinthetaT-1/2gT^2