1) Consider water flowing out a reservoir through a tube as shown. The level of
ID: 1352431 • Letter: 1
Question
1) Consider water flowing out a reservoir through a tube as shown. The level of the water in the reservoir is a height, h, above the tube's opening. You may assume that the area of the reservoir is large compared to the area of the tube, thus the level in the reservoir is changing at a negligbly small velocity.
a) Using Bernoulli's equation, find the velocity with which the water leaves the tube.
b) Does your answer depend on the area of the tube?
2) Consider a ball thrown horizontally off the top of a building of height y. Some time later the ball will hit the ground a horizontal distance x from where it was thrown. In terms of x and y, what was the inital speed of the ball.
A= cross sectional areaExplanation / Answer
bernoulli's equation states that,
P+D*g*h+0.5*D*v^2=constant
where P=pressure
D=density of water
g=9.8 m/s^2
h=height
v=speed
a)initial condition:
P1=atmospheric pressure=P0
v=0 m/s
height=h
final condition:
v=?(to be determined)
P2=atmospheric pressure=P0
height=0
then using the bernoulli's equation:
P0+D*g*h+0.5*D*0^2=P0+D*g*0+0.5*D*v^2
==>0.5*D*v^2=D*g*h
==>v=sqrt(2*g*h)
b) no, the answer does not depend upon area of the tube
Q2.
initial vertical speed=0 m/s
vertical acceleration=9.8 m/s^2
distance covered=y m
let time taken to reach ground is t seconds.
then using the formula:
distance=initial speed*time+0.5*acceleration*time^2
==>y=0*t+0.5*9.8*t^2
==>t=sqrt(y/4.9) seconds
if initial horizontal speed is v,
then horizontal speed=v*t
=v*sqrt(y/4.9)
hence x=v*sqrt(y/4.9)
==>v=x/sqrt(y/4.9)
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