Answer
See the detailed answer below.
Work Step by Step
a) To find the speed with which the water exits the hole at a height $y$, we need to use Bernoulli’s equation (since we assumed that the water is an ideal fluid) at the two points at height $h$ and at height $y$.
Let's say that at height $h$ is point 1 and at height, $y$ is point 2.
$$\color{red}{\bf\not}P_1+\frac{1}{2}\rho v_1^2+\rho gy_1=\color{red}{\bf\not}P_2+\frac{1}{2}\rho v_2^2+\rho gy_2$$
Noting that at $y_1=h$ and $y_2=y$, the pressure is the atmospheric pressure. So, $P_1=P_2$.
$$ \frac{1}{2}\rho v_1^2+ \rho gy_1= \frac{1}{2}\rho v_2^2+ \rho gy_2$$
The speed of the water at $y=h$ is zero because the water is continuously replenished.
$$ \frac{1}{2}\rho (0)^2+\rho gy_1= \frac{1}{2}\rho v_2^2+\rho gy_2$$
$$ \color{red}{\bf\not}\rho gh= \frac{1}{2}\color{red}{\bf\not}\rho v_2^2+ \color{red}{\bf\not}\rho gy$$
Solving for $v_2$;
$$\boxed{v_2=\sqrt{ 2g(h-y) }}$$
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b) Since there are no forces acting on the object horizontally, and the only force acting on it vertically is its own weight, we can use the kinematic formulas to find its range.
First, we need to find its trip from the height $y$ to the ground and this object has zero initial $y$-velocity component.
$$y_f=y_i+v_{iy}t+\frac{1}{2}a_yt^2$$
Hence,
$$0=y+0+\frac{1}{2}(-g)t^2$$
Hence,
$$t=\sqrt{\dfrac{2y}{g}}\tag 1$$
The range is given by
$$x=x_i+v_{ix}t+\frac{1}{2}a_xt^2$$
$$x=0+v_{ix}t+0$$
Plugging from (1)
$$\boxed{x=v\sqrt{\dfrac{2y}{g}}}$$
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c) Plugging $v$ from the first boxed formula in part (a) above, where $v_2$ is horizontal.
$$x=\sqrt{ 2\color{red}{\bf\not}g(h-y) }\sqrt{\dfrac{2y}{\color{red}{\bf\not}g}}$$
$$x=\sqrt{4y( h-y )}\tag 2$$
To find the maximum range we need to differentiate creative to $y$ at which $dx/dy=0$
Hence,
$$\dfrac{dx}{dy}=\dfrac{d }{dy}\left[4y( h-y )\right]^{\frac{1}{2}}=\dfrac{d }{dy}\left[2(y^{\frac{1}{2}})( h-y )^{\frac{1}{2}}\right] =0$$
$$ y^{-\frac{1}{2}} ( h-y )^{\frac{1}{2}} + \frac{1}{2}(h-y)^{-\frac{1}{2}} (-1)2y^{ \frac{1}{2}}=0$$
$$\dfrac{\sqrt{h-y}}{\sqrt{y}}- \dfrac{\sqrt{y}}{\sqrt{h-y}}=0$$
Hence,
$$\dfrac{\sqrt{h-y}}{\sqrt{y}}= \dfrac{\sqrt{y}}{\sqrt{h-y}}=0$$
$$y=h-y$$
$$2y=h$$
Thus,
$$\boxed{y_{max}=\dfrac{h}{2}}$$
Therefore the maximum range is given by plugging $y_{max}$ into (2);
$$x=\sqrt{(4)\dfrac{h}{2}\left( h-\dfrac{h}{2}\right)} =\sqrt{2h^2-h^2}=h$$
$$\boxed{x_{max}=h}$$
