Water stands at a depth H in a large open tank whose side
walls are vertical (Fig. 3-15). A hole is made in one of the walls at a
depth h below the water surface.
(a) At what distance R from the foot of the wall does the
emerging stream of water strike the floor?
(b) At what height above the bottom of the tank could a
second hole be cut so that the stream emerging from it would have the same range?
Answers
Explanation:
The velocity is given as
v=
2gh
Let the hole was made at heighthbelow the water level
Then
H−h=
2
1
gt
2
By using the equation
v=
t
d
R=v×t
By substituting the time from the above equation we get
R=
2gh
×(
g
2(H−h)
)
2
1
On solving the above equation
R=2
h(H−h)
The maximum range can be found by
dh
dR
=0
Hence the maximum range isH=2h
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