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Physics- How far away do you have to hold the glass so that the wine stream hits it exactly?

You are visiting a winery in Llano, Texas, for a wine tasting. A large storage barrel (diameter of 4.2) is filled to a height of 260 cm. The barrel has on its top a small hole to guarantee contact with the atmospheric pressure. The tap of the barrel is at a height of 15 cm from the bottom of the barrel. The opening of the tap faces towards you and does not point downward.

if you read my explanation to one of your questions last time, you would have been able to do this.you didnt give me a best answer when i explained the whole thing for you without numbers.anyways,i guess u want to see numbers.fine then,

1-bernoullis equation for the top small hole and the opening.point 1 is top, point 2 is opening.

P1 + Dgh1 +1/2Dv1^2 = P2 + Dgh2 +1/2Dv2^2

key: P1=P2=P atmosphere , v1=velocity 1 = ~0 m/s

so: Dgh1 =Dgh2 +1/2Dv2^2
rearrange: SQRT [ 2g( h1- h2 ) ] = v2 (velocity of water coming out of tap)

v2= SQRT [ 2*9.81m/s^2 * ( 2.6m - 0.15 m) ] = 6.93 m/s =v2

2-use horizontally launched projectile motion equations:
V in x direction = constant, Vx= dx/dt (d is delta) , so
delta t= delta x/Vx , where v2 from before=Vx = 6.93 m/s

another kinematics equation is :
delta y = Viy* delta t – 1/2*g*(delta t) ^2 . Viy=V intial y-direction=0m/s
delta y= – 1/2*g*(delta t) ^2
now subsitute delta t with delta x/ Vx
delta y= -1/2 *g* (delta x/Vx )^2
now solve for delta x , because you know delta y= – 0.15m (because going down, its negative)
-0.15m = -1/2 *g* (delta x/Vx )^2
Vx * SQRT [ 0.3m/9.81m/s^2 ] = delta x
6.93 m/s * 0.1749s =delta x
1.212 m =delta x
delta x = 121.2 cm
so the WINEGLASS HAS TO BE EXACTLY 121.2 cm away from the tap

Auto Vino Wine Storage Menlo Park presents Virtual Wine Tip Clip- Buying a Wine Glass.mp4