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

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

$v = \sqrt{2gh}$  

for upper hole  $v = \sqrt{2\times 9.8 \times 19}$  

                         v  = 19.29 m/s

lower hole    $v = \sqrt{2\times 9.8 \times 117}$  

                          v = 47.88 m/s

The path for each is parabolic

x = v t

$y = \dfrac{1}{2}gt^2$  

$y = \dfrac{1}{2}g(\dfrac{x}{v})^2$  

$y = \dfrac{gx^2}{2v^2}$  

we get

upper hole

$y = \dfrac{9.8\times x^2}{2\times 19.29^2}= 0.0132 x^2$  lower hole

$y= \dfrac{9.8\times x^2}{2\times 47.88^2}=0.00214x^2$  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