Solution: Sprinkler Range & Area
When the sprinkler is on the ground, the maximum vertical height is $h$. Using kinematics $v^2 = u^2 + 2as$ (with final velocity 0): $$ 0 = v^2 – 2gh \implies v = \sqrt{2gh} $$
Why is the boundary equation $z = \frac{v^2}{2g} – \frac{gx^2}{2v^2}$?
The equation of trajectory for a projectile launched at angle $\theta$ is:
$$ z = x \tan \theta – \frac{g x^2}{2v^2 \cos^2 \theta} $$
Using the identity $\frac{1}{\cos^2 \theta} = 1 + \tan^2 \theta$, let $m = \tan \theta$:
$$ z = xm – \frac{g x^2}{2v^2}(1 + m^2) $$
Rearranging this as a quadratic equation in terms of $m$:
$$ \left(\frac{g x^2}{2v^2}\right)m^2 – (x)m + \left(z + \frac{g x^2}{2v^2}\right) = 0 $$
For the boundary (the maximum possible reach), this quadratic must have exactly one solution for $\tan \theta$ (tangent to the envelope). Thus, the discriminant $\Delta$ must be zero:
$$ \Delta = B^2 – 4AC = (-x)^2 – 4\left(\frac{g x^2}{2v^2}\right)\left(z + \frac{g x^2}{2v^2}\right) = 0 $$
$$ x^2 – \frac{2g x^2}{v^2}\left(z + \frac{g x^2}{2v^2}\right) = 0 $$
Dividing by $x^2$ and simplifying:
$$ 1 – \frac{2g}{v^2}\left(z + \frac{g x^2}{2v^2}\right) = 0 \implies \frac{v^2}{2g} = z + \frac{g x^2}{2v^2} $$
$$ \therefore z = \frac{v^2}{2g} – \frac{g x^2}{2v^2} $$
The sprinkler is now at height $H=h$. We set the nozzle as the origin $(0,0)$. The ground is at $z = -h$.
Substitute $z = -h$ and $v^2 = 2gh$ into the envelope equation:
$$ -h = \frac{(2gh)}{2g} – \frac{g R^2}{2(2gh)} $$
$$ -h = h – \frac{g R^2}{4gh} $$
$$ -2h = – \frac{R^2}{4h} \implies R^2 = 8h^2 $$
New Area $A_2 = \pi R^2 = 8 \pi h^2$.
Original Area (on ground) had radius $R_1 = 2h$. So $A_1 = \pi (2h)^2 = 4 \pi h^2$.
The ratio is $\frac{8 \pi h^2}{4 \pi h^2} = 2$.
This method uses the logic that the envelope is a parabola.
The envelope is a parabola with its vertex at height $h$ above the nozzle (since $v^2/2g = h$). For a parabola centered at the vertex, the square of the horizontal radius is proportional to the vertical distance from the vertex ($R^2 \propto \Delta z$).
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At Nozzle Level ($z=0$):
The vertical distance from the vertex is $h$. Let the radius here be $R_{nozzle}$. -
At Ground Level ($z=-h$):
The vertical distance from the vertex is $h + h = 2h$. Let the radius here be $R_{ground}$.
Since the vertical distance doubles ($h \rightarrow 2h$), the square of the radius ($R^2$) must also double. $$ R_{ground}^2 = 2 \times R_{nozzle}^2 $$ Since Area $A = \pi R^2$, the area also strictly doubles. $$ A_{ground} = 2 \times A_{nozzle} $$