A vessel is filled with a liquid of density 1900 kg/m3 . There are two holes (one above the other) in the side of the vessel. Liquid streams out of both of these holes in a horizontal direction. The upper hole is a distance 19 m below the surface of the liquid and the lower hole is a distance 117 m below the surface of the liquid. The acceleration of gravity is 9.8 m/s 2 . At what horizontal distance from the vessel does the stream from the upper hole collide with the stream from the lower hole?

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wagonbelleville wagonbelleville · Answer 1 · 2019-10-11T07:57:21+02:00

Answer:

110 meters is the distance where they will intersect

Explanation:

given,

liquid density = 1900 kg/m³

distance of upper hole = 19 m

distance of lower hole = 117 m

acceleration due to gravity = 9.8 m/s²

the speed at each point

[tex]v = \sqrt{2gh}[/tex]

for upper hole [tex]v = \sqrt{2\times 9.8 \times 19}[/tex]

v = 19.29 m/s

lower hole [tex]v = \sqrt{2\times 9.8 \times 117}[/tex]

v = 47.88 m/s

The path for each is parabolic

x = v t

[tex]y = \dfrac{1}{2}gt^2[/tex]

[tex]y = \dfrac{1}{2}g(\dfrac{x}{v})^2[/tex]

[tex]y = \dfrac{gx^2}{2v^2}[/tex]

we get

upper hole

[tex]y = \dfrac{9.8\times x^2}{2\times 19.29^2}= 0.0132 x^2[/tex] lower hole

[tex]y= \dfrac{9.8\times x^2}{2\times 47.88^2}=0.00214x^2[/tex] y for upper hole = 80 + y for lower hole

[tex]0.0132 x^2= 98 + 0.00214 x^2[/tex]

[tex]0.0082 x^2 = 98 [/tex]

x = 109.32 meters

110 meters is the distance where they will intersect