Question

A jet of water squirts out horizontally from a hole near the bottom of the tank, as seen in the figure.

a) Assume that x = 1.48 m and y = 1.85 m. What is the speed of the water coming out of the hole?

b) If the hole has a diameter of 3.97 mm, then what is the height, h, of the water level in the tank?

Answer 1

Given that

The horizontal distance (x) =1.48m

The vertical height (y)=1.85m

The diameterof the hole (d) =3.97mm =0.00397m

Acceleration due to gravity (g) =9.81m/s2

The time for which the water coming from the hole is given by

s =ut+(1/2)at²

y =(1/2)at²

time (t) =Sqrt(2y/a) =Sqrt(2*1.85/9,81) =0.614s

Now horizontal exit velocity of the water from the hole is (vx) =x/t =1.48m/0.614s=2.410m/s

b)

Now from the Bernoullis law at the bottom and tank surface level

we have

P + (1/2)ρv₀² + ρgh = P + (1/2)ρv_x² + ρgh

P + (1/2)ρ(0)² + ρgh = P + (1/2)ρv_x² + ρg(0m)

gh =(1/2)v_x²

h =v_x²/2g =(2.410m/s)²/2*9.81 =0.296m or 29.60cm