Antigravity in the summertime

Velocity of water at the maximum height equals

$\displaystyle v^2 = v_0^2 - 2gh_{max}$ ....... (1)

Now force provided by water stream should balance the weight of tray

Change in momemtum of stream = Mg

$J( v - (-v)) = Mg$

$\displaystyle v = \dfrac{Mg}{2J}$

Substituting in equation (1)

$\displaystyle \left( \dfrac{Mg}{2J} \right)^2 = v_0^2 - 2gh_{max}$

Substituting given values in the equation to get

$\displaystyle \boxed{ h_{max} = 6.797m}$

To keep the tray floating at rest, net force on tray must be zero. The weight of the tray is $W _t= Mg$ downwards. The water must provide the same magnitude of force upwards.

If water collides elastically with the tray, then the momentum of water changes at the rate of $J_m (v-(-v)) = 2J_m v$ . Using Newton's third law of motion, water will exert equal and opposite force on tray i.e. $2 J_m v$ force upward. Here $v$ is the velocity at which water collides with tray.

Both forces are equal, so

$Mg = 2 J_m v$

$\implies v = \dfrac{Mg}{2 J_m} \phantom{.......} (*)$

Using laws of kinematics,

$v^2 - v_{0}^{2} = 2(-g) h_{\text{max}}$

$\implies v = \sqrt{v_{0}^{2} - 2gh_{\text{max}}} \phantom{.......} (**)$

We use $(*)$ and $(**)$ to eliminate $v$

$\dfrac{Mg}{2 J_m} = \sqrt{v_{0}^{2} - 2gh_{\text{max}}}$

We make $h_{\text{max}}$ the subject of the formula.

$h_{\text{max}} = \frac{v_{0}^{2} - \left(\frac{Mg}{2 J_m}\right)^2}{2g}$

When we substitute the values, we get $h_{\text{max}} \approx \boxed{6.797\text{m}}$ $_\square$