Answer:
See description
Explanation:
This is an example where we need Tornicelli's law, which states that the horizontal speed of a fluid that starts falling from an orifice is the same speed that an object acquires from free-falling.
[tex]v = \sqrt{2gh}[/tex]
we are given:
[tex]h_{cilinder} = 0.2 [m]\\h = 0.05 [m]\\d=0.15[m][/tex]
the horizontal velocity of the water at the start is:
[tex]v = \sqrt{2(9.8)(0.05)}=0.989949 [m/s]=1[m/s][/tex]
now we need to find the time for the water drops to fall d:
as the gravity is the only force interacting with the water we have:
[tex]y(t) = \frac{1}{2} g*t^2[/tex]
replace for y = d
[tex]0.15 = \frac{1}{2} g*t^2=>t=\sqrt{\frac{2*0.15}{9.8}}=0.1749[s] [/tex]
now that we have t we notice that there are no horizontal forces interacting with the water, so the horizontal position is given by:
[tex]x(t)=v*t[/tex]
Finally, we replace v and t:
[tex]x(2.45) = 1*0.1749 = 0.1749 [m]=17.49[cm][/tex]
