I'm not really sure how to do this Projectile Motion question since it gives no equations of motion or anything, just a lot of text and conditions:

A tall building stands on level ground. The nozzle of a water sprinkler is positioned at a point $P$ on the ground at a distance $d$ from a wall of the building. Water sprays from the nozzle with speed $V$ and the nozzle can be pointed in any direction from $P$.

Suppose that $V=\sqrt{2gd}$. Show that the portion of the wall that can be sprayed with water is a parabolic segment of height ${8\over 15}d$ and area ${5\over 2}d^2\sqrt{15}$.

Someone has already asked this question, but they solved it before anyone else could answer it, so I was wondering if anyone else would be able to assist me? I tried to solve this question by setting the angle in the vertical plane to a constant $\frac{\pi}{4}$ for max range, and varied the angle of the diagonal in which the trajectory of the projectile would lie in. I ended up getting a result:

$y=\sqrt{x^2+d^2} - {x^2\over 4d}$

I'm certain this isn't a parabola though so I wasn't sure what I did wrong. Could someone point out the flaw in this argument? For reference, the equation of the parabola is, I believe:

$y=-{x^2\over 8d} + {15\over 8}d$


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    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Aug 4 '17 at 15:18
  • $\begingroup$ I fixed it. Could you unhold this post now, please? $\endgroup$ – Galaxy Nova Aug 5 '17 at 0:27
  • $\begingroup$ You're unlikely to have it reopened since it lacks any evidence that you've tries the problem yourself nor have you given any context so that we can answer at a level suitable for you. I suggest that you edit the question accordingly. Don't be deterred though: I'll upvote it once it's done that way. $\endgroup$ – Shaun Aug 5 '17 at 0:32
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    $\begingroup$ Is it okay now? $\endgroup$ – Galaxy Nova Aug 5 '17 at 2:05
  • $\begingroup$ You shouldn’t be looking for maximum range, but maximum height at a given distance. $\endgroup$ – amd Aug 5 '17 at 6:22

There seems to be a typo in the problem : The $V = 2\sqrt{gd}$ and then the rest follows as given below in the solution. I think I got what the book answer is.

Solution as follows:

enter image description here enter image description here

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    $\begingroup$ Yes, looks like it should be $V=2\sqrt{gd}$. $\endgroup$ – Hypergeometricx Aug 6 '17 at 7:13
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    $\begingroup$ I wasted a lot of time on the OP's incorrect expression. That is OK, I have become a master now trying to figure out how it could work. $\endgroup$ – Satish Ramanathan Aug 6 '17 at 7:21

$\hspace{4.5cm}$enter image description here

General projectile trajectory equation: $$y=u\tan\alpha-\frac {gu^2}{2V^2}\sec^2\alpha$$ At $u=\ell$, $$\begin{align} y&=\ell \tan\alpha-\frac {g\ell^2}{2V^2}\sec^2\alpha\\ \frac{\text dy}{\text d\alpha}&=\ell\sec^2\alpha-\frac {g\ell^2}{2V^2}\cdot 2\sec\alpha\cdot \sec\alpha\tan\alpha\\ &=\ell\sec^2\alpha\left(1-\frac {g\ell^2}{V^2}\tan\alpha\right) =0\;\text{when}\; \tan\alpha=\frac {V^2}{g\ell}\\ \therefore y_\text{max}=h&=\ell\left(\frac {V^2}{g\ell}\right)-\frac {g\ell^2}{2V^2}\left(1+\frac {V^4}{g^2\ell^2}\right)\\ &=\frac {V^2}{2g}-\frac {g\ell^2}{2V^2}\end{align}$$ For a given lateral displacement $x$, $\ell^2=d^2+x^2$ $$\therefore h=\frac {V^2}{2g}-\frac g{2V^2}(d^2+x^2)$$ Given $V=2\sqrt{gd}$ i.e. $V^2=4gd$, hence $$h=2d-\frac 1{8d}(d^2+x^2)\\ =\color{red}{\frac {15}8 d-\frac {x^2}{8d}}$$

From the equation, when $x=0$, the height at the centre, i.e. maximum height is given by $$\color{red}{h=\frac {15}8d}$$. Area under the curve is given by

Also, when $y=0$, $x=\pm d\sqrt{15}$. As the curve is symmetrical, the area under the curve is given by $$\begin{align} 2 \int_0^{d\sqrt{15}}\left(\frac {15}8d-\frac {x^2}{8d} \; \right)\text dx &=2\cdot \frac 18 \left[15d\cdot x-\frac {x^3}{3d}\right]_0^{d\sqrt{15}}\\ &=\color{red}{\frac 52d^2\sqrt{15}} \end{align}$$

It is interesting to note that $$\alpha=\arctan\left(\frac {4d}{\sqrt{d^2+x^2}}\right)$$ i.e. $\alpha$ decreases slowly with increasing $x$, meaning that as the projectile moves away from the centre (plane perpendicular to wall) the angle of launch must be decreased slightly to achieve maximum height on hitting the wall.

NB: here we use $y$ for general height of projectile and $h$ for maximum height when it hits the wall.

ADDENDUM - Another Method

Using a cylindrical coordinate system $(r,\theta, y)$ with radius $r$, angle $\theta$ and vertical distance $y$, the equation for general family of trajectories launched from the origin (i.e. $P$) in direction $\theta$ and at angle $\alpha$ from the plane $y=0$ is given by $$\begin{align} y &=r\tan\alpha-\frac {gr^2}{2V^2}\sec^2\alpha\\ &=r\left(\tan\alpha-\frac r{8d}\sec^2\alpha\right)&&(V=2\sqrt{gd})\tag{1} \end{align}$$

The plane of the wall is given by $$r=\frac d{\sin\theta}\tag{2}$$ Putting $(2)$ in $(1)$ gives $$\begin{align} y &=\frac d{\sin\theta}\left(\tan\alpha-\frac {\sec^2\alpha}{8\sin\theta}\right)\\ \frac {\text dy}{\text d\alpha} &=\frac {d\cdot \sec^2\alpha}{\sin\theta}\left(1-\frac {\tan\alpha}{4\sin\theta}\right)\\ &=0\;\text{when}\; \tan\alpha=4\sin\theta\\ \therefore y^* &=\frac d{\sin\theta}\left(4\sin\theta-\frac {1+16\sin^2\theta}{8\sin\theta}\right)\\ &=d\left(2-\frac 1{8\sin^2\theta}\right)\\ &=2\left(2-\frac {d^2+x^2}{8d^2}\right)\\ &=\color{red}{\frac {15d}8-\frac {x^2}{8d}} \end{align}$$

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