Question

Water stands at a depth H in a tank, Whose side walls are vertical. A hole is made on one of the walls at a depth h below the water surface. Find at what value of h this range is maximum?

Answer 1

Condition:

Water stands at a depth $\mathrm{H}$ in a tank, Whose side walls are vertical. A hole is made on one of the walls at a depth $\mathrm{h}$ below the water surface. Find at what value of $\mathrm{h}$ this range is maximum?

Solution:

Apply Bernoulli's equation for water located inside the container at depth $h$ and for water just flowing out of it at the same depth: $p_{atm} + \rho gh = p_{atm} + \frac{\rho v^2}{2}$ . We obtain $v = (2gh)^{0.5}$ .

Here we have a projectile problem that starts horizontally at height $H - h$ with the speed $v$ .

It accelerates downward with free fall acceleration, and the travelling time until the water hits the ground is found from the condition $H - h = \frac{gt^2}{2}$ , therefore $t = \left(\frac{2(H - h)}{g}\right)^{0.5}$ . During that time interval the horizontal distance travelled is $vt$

$$x = v t = (2 g h) ^ {0. 5} * \left(\frac {2 (H - h)}{g}\right) ^ {0. 5} = \left(\frac {2 (H - h) 2 g h}{g}\right) ^ {0. 5} = 2 ((H - h) h) ^ {0. 5}.$$

For maximum range:

$$\frac {d x}{d h} = \frac {d}{d h} \left(2 \left((H - h) h\right) ^ {0. 5}\right) = 0 \rightarrow H - 2 h = 0 \rightarrow h = \frac {1}{2} H.$$

Answer: $\frac{1}{2} H$ .