Bernoulli's energy conservation principle for the fluids tells the relation between pressures and kinetic energy at two locations in a fluid.
Total Energy at the surface of water = Total energy at the hole at 20 cm depth
P₁ + 1/2 ρ v₁² + ρ g h₁ = P₂ + 1/2 ρ v₂² + ρ g h₂
P₁ = P₂ = atmospheric pressure, g = 10 m/sec²
v₁ = 0 as the water at the surface is maintained stationary.
h₁ - h₂ = h = 20 cm = 0.20 m
So v₂ = √ (2gh) --- equation 1
Hence, v₂² = 2 * 0.20 g = 4 => v₂ = 2 m/sec
We find the range for the water particles with this speed horizontally directed and at a height of 0.40 meters above ground.
t = time to reach ground = √ (2 * 0.40 / g ) = 0.2 *√2 sec
Range = v₂ t = 0.4 √2 meters
If h is the depth of hole from water surface, then v = √(2gh)
time to reach ground = t = √ [2*(0.60-h)/g ]
Then range R = v₂ t = 2 √[ h (60-h) ] = 0.4√2 --- equation 2
Range will be equal for h = 20cm, and h = 40 cm, as the product of h and (60-h) will be same. If you solve quadratic equation 2, we get the same.
velocity v₂ of water coming out of hole at the bottom of tank = √(2 g h) from equation 1 above. Here we neglect v₁ at the water surface as it is very small as compared to v₂. Let A be the area of cross section of the tank.
Volume of water in tank = V = A h
water flow rate out of tank = decrease rate in volume inside tank
A v₂ = d V/dt = - A dh/dt
=> v₂ = √(2 g h ) = - dh/dt
=> - dh/√h = √(2g) dt ------- equation 3
Integrating the expressions on both sides ,
Time for water level to go from h₀ to 3 h₀/4, is
Time for water level to go from 3 h₀ /4 to 0 is
Ratio t1/t2 =
3) Using equation 4, above we get the values for t1 and t2.