Question

a fountain shoots water out at an angle of 70 degrees above the horizontal. if the maximum height of the water stream is 11.5cm, what is the initial velocity of the water as it leaves the spigot? How do you find it when they give you this less info?

Answer 1

I know the answer's 1.6m/s but how do you get that answer? help please.

Answer 2

Can anyone help me???

Answer 3

Hold please...

Answer 4

Knowing that Vy=o at the top of the stream. Solve Vyfinal =\[\sqrt{2*a*d}\]

Answer 5

=\[\sqrt{2*9.81*.115m}\]

Answer 6

=1.5 m/s

Answer 7

Now Vy=V*sin70

Answer 8

Then, V=Vy/sin70 = 1.6 m/s is the answer.

Answer 9

There is a formula for the max height which can be easily derived.

\(\huge{h_{max} = \frac{ v^{2} sin^{2} \theta}{2g}}\)

We know \(\theta\) =70 and \(h_{max}\) =0.115m

You can now find the velocity,v with which the particle was projected.

Answer 10

the formula Diwakar wrote comes from this...
\[v^2=u^2-2gh\]
where v=final velocity
u=initial velocity
g=gravitational acceleration
h=height

at maximum height h=\(h _{\max}\)
v=0,u=\(usin\theta\)
so we get
\(\huge{h _{\max}=\frac{ u^2\sin^2 \theta }{ 2g }}\)

Answer 11

Hey dfav how come when I \[\sqrt{2*9.81*115}\] I got the answer 47.5 m/s. What did I do wrong here? I multiplied them which was 2254 and when I square rooted it gave me the number 47.47m/s.

Answer 12

\[u=\frac{ \sqrt{h _{\max }*2g} }{ \sin \theta }\]

Answer 13

Late reply...the units for velocity are m/s. So when calculting \[\sqrt{2*a*d}\], d = 11.5 cm must be converted to meters.
Now, d=0.115 m.
\[Then, V = \sqrt{2*9.81 m/s^2*0.115 m/s} = 1.50 m/s\]