Question

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?

Answer 1

110 meters is the distance where they will intersect

Step-by-step 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

`v = √(2gh)`

for upper hole
`v = √(2* 9.8 * 19)`

v = 19.29 m/s

lower hole
`v = √(2* 9.8 * 117)`

v = 47.88 m/s

The path for each is parabolic

x = v t

`y = (1)/(2)gt^2`

`y = (1)/(2)g((x)/(v))^2`

`y = (gx^2)/(2v^2)`

we get

upper hole

`y = (9.8* x^2)/(2* 19.29^2)= 0.0132 x^2` lower hole

`y= (9.8* x^2)/(2* 47.88^2)=0.00214x^2` y for upper hole = 80 + y for lower hole

`0.0132 x^2= 98 + 0.00214 x^2`

`0.0082 x^2 = 98`

x = 109.32 meters

110 meters is the distance where they will intersect