Toricelli's Law of Fluid Flow

• (a) Show that the standard gravity differential equation $\frac{d^2h}{dt^2} = -g$ implies that an object dropped from height $h(0)$ lands with velocity $-\sqrt{2gh(0)}$.

  • We solve this by integrating:

    $\frac{dh}{dt} = -gt + C_1$
    $h(t) = -\frac{1}{2}gt^2 + C_1 t + C_2$

    Apply the initial condition that the object is dropped:

    $v(0)=0 \Rightarrow C_1 = 0$

    So:

    $\frac{dh}{dt} = -gt$
    $h(t) = h_0 - \frac{1}{2}gt^2$

    Solve for $t$ and substitute back:

    $v = -\sqrt{2g(h_0 - h)}$
• (b)
• (c) The conical tank has a radius of 20 cm when it's filled to an initial depth of 50cm. A small round hole at the bottom has a diameter of 1cm. Determine A(h) and a then solve the differential equation $A(h)\frac{dh}{dt} = -a * \sqrt{2*g*h}$ and relate time and the height if the water.
  Let's do integration $\int{A(h)*dh/dt = \int{-a*\sqrt{2*g*h}}}$ and then