Physicspupil – Page 31

May 24, 2021October 24, 2022

202105241107 Homework 1 (Q2)

Prove that the divergence of a curl is always zero; and the curl of a gradient is always zero.

Solution.

First,

Let $\mathbf{A}=A_x\,\hat{\mathbf{i}} + A_y\,\hat{\mathbf{j}} + A_z\,\hat{\mathbf{k}}$ .

$\begin{aligned} & \quad \nabla \cdot (\nabla\times \mathbf{A}) \\ & = \nabla \cdot \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} \\ & = \nabla \cdot \Bigg( \bigg( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \bigg)\,\hat{\mathbf{i}} - \bigg( \frac{\partial A_z}{\partial x} - \frac{\partial A_x}{\partial z} \bigg)\,\hat{\mathbf{j}} + \bigg( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \bigg)\,\hat{\mathbf{k}} \Bigg) \\ & = \frac{\partial^2 A_z}{\partial x\partial y} - \frac{\partial^2 A_y}{\partial x\partial z}-\frac{\partial^2 A_z}{\partial y\partial x} + \frac{\partial^2A_x}{\partial y\partial z} + \frac{\partial^2A_y}{\partial z\partial x}-\frac{\partial^2A_x}{\partial z\partial y} \\ & = 0 \end{aligned}$

$\therefore$ The divergence of a curl is always zero.

Second,

$\nabla f(x,y,z) = \displaystyle{\bigg( \frac{\partial f(x,y,z)}{\partial x} ,\enspace \frac{\partial f(x,y,z)}{\partial y},\enspace \frac{\partial f(x,y,z)}{\partial z}\bigg)}$
$\mathbf{F}\stackrel{\textrm{def}}{=} \nabla f$

$\begin{aligned} \nabla \times \mathbf{F} & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{vmatrix} \\ & =\bigg( \frac{\partial^2f}{\partial y\partial z} - \frac{\partial^2f}{\partial z\partial y},\enspace \frac{\partial^2f}{\partial z\partial x} - \frac{\partial^2f}{\partial x\partial z},\enspace \frac{\partial^2f}{\partial x\partial y}-\frac{\partial^2f}{\partial y\partial x} \bigg) \\ & = 0 \\ \end{aligned}$

$\therefore$ The curl of a gradient is always zero.

May 24, 2021October 24, 2022

202105241031 Homework 1 (Q1)

Let $\boldsymbol{\lambda}\lambda$ be the separation vector from a fixed point $(x',y',z')$ to the point $(x,y,z)$ , and let $\lambda$ be its length. Show that

(a) $\nabla (\lambda^2)=2\boldsymbol{\lambda}$

(b) $\nabla (1/\lambda )=-\hat{\boldsymbol{\lambda}}/\lambda^2$

(c) What is the general formula for $\nabla (\lambda^n)$ .

Solution.

(a)

$\begin{aligned} \boldsymbol{\lambda}\lambda & = (x-x',y-y',z-z') \\ \lambda & = \textrm{length } = \sqrt{(x-x')^2+(y-y')^2+(z-z')^2} \\ \lambda^2 & = (x-x')^2 + (y-y')^2 + (z-z')^2 \\ \nabla (\lambda^2)& =\Big( 2(x-x'), 2(y-y'), 2(z-z')\Big) \\ & = 2\boldsymbol{\lambda} \\ \end{aligned}$

(b)

$\begin{aligned} & \quad \nabla \bigg( \frac{1}{\lambda} \bigg) \\ & = \nabla \bigg( \Big( (x-x')^2 + (y-y')^2 + (z-z')^2 \Big)^{-\frac{1}{2}} \bigg) \\ & = -\frac{1}{2} \Big( (x-x')^2 + (y-y')^2 + (z-z')^2 \Big)^{-\frac{3}{2}} \cdot \nabla (\lambda^2) \\ & \stackrel{\textrm{(a)}}{=} -\frac{1}{2}(\lambda^2)^{-\frac{3}{2}}\cdot 2\boldsymbol{\lambda} \\ & = -\lambda^{-3}\cdot \lambda\hat{\boldsymbol{\lambda}} \\ & = -\frac{\hat{\boldsymbol{\lambda}}}{\lambda^2} \\ \end{aligned}$

(c)

$\begin{aligned} \nabla (\lambda^n) & = \nabla (\lambda^2)^{\frac{n}{2}} \\ & = \frac{n}{2}(\lambda^2)^{\frac{n}{2}-1}\cdot\nabla (\lambda^2) \\ & \stackrel{\textrm{(a)}}{=} \frac{n}{2}(\lambda^2)^{\frac{n-2}{2}}\cdot 2\boldsymbol{\lambda} \\ & = \frac{n}{2}(\lambda^2)^{\frac{n-2}{2}}\cdot 2\lambda\hat{\boldsymbol{\lambda}} \\ & = n\lambda^{n-1}\hat{\boldsymbol{\lambda}} \\ \end{aligned}$

April 24, 2021October 24, 2022

202104241713 Homework 1 (Q7)

The height of a mountain is given by $h(x,y)=3000-2x^2-y^2$ , where the $y$ -axis points east, the $x$ -axis points north, and all distances are measured in meters. Suppose a mountain climber is at the point $(30,\, -20,\, 800)$ , will he ascend or descend if he moves in the southwest direction?

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2155 Methods of Physics II Homework Solutions.)

The altitude $h(x,y)$ is given by a function of $x$ and $y$ :

$h(x,y)=3000-2x^2-y^2$ .

Now that the climber moves in the southwest direction

$\begin{aligned} \mathbf{n} & =-1\,\hat{\mathbf{i}}-1\,\hat{\mathbf{j}} \\ \hat{\mathbf{n}} & = \frac{1}{\sqrt{2}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \end{aligned}$

$\begin{aligned} h_{\hat{\mathbf{n}}}'(x,y) & = \nabla h(x,y)\cdot \hat{\mathbf{n}} \\ & = (-4x\,\hat{\mathbf{i}}-2y\,\hat{\mathbf{j}}) \cdot \frac{1}{\sqrt{2}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \\ & = \frac{1}{\sqrt{2}} (4x+2y) \end{aligned}$

At point $(30,\, -20,\, 800)$ ,

$\begin{aligned} h_{\hat{\mathbf{n}}}'(30,-20) & = \frac{1}{\sqrt{2}}\big( 4(30)+2(-20)\big) \\ & = \frac{1}{\sqrt{2}}\cdot 80 \qquad (>0) \end{aligned}$

he will ascend southwesterly.

April 22, 2021October 24, 2022

202104221623 Homework 2 (Q1)

Find the gradients of the following functions:

(a) $\phi (x,y,z)=x^2+y^2+z^2$ ;

(b) $\phi (x,y,z)=\exp (x+y)\cos z$ .

Solution.

(a)

$\begin{aligned} & \quad \nabla \phi \\ & = \bigg(\hat{\mathbf{i}}\frac{\partial}{\partial x}+\hat{\mathbf{j}}\frac{\partial}{\partial y}+\hat{\mathbf{k}}\frac{\partial}{\partial z}\bigg)\phi (x,y,z) \\ & = \hat{\mathbf{i}} \frac{\partial}{\partial x}(x^2+y^2+z^2) + \hat{\mathbf{j}} \frac{\partial}{\partial y}(x^2+y^2+z^2) + \hat{\mathbf{k}} \frac{\partial}{\partial z}(x^2+y^2+z^2) \\ & = 2x\,\hat{\mathbf{i}} + 2y\,\hat{\mathbf{j}} + 2z\,\hat{\mathbf{k}} \end{aligned}$

(b) It is left as an exercise to the reader.

April 22, 2021November 25, 2024

202104221542 Homework 2 (Q4)

Evaluate the limit, or explain why the limit fails to exist.

(a) $\displaystyle{\lim_{(x,y)\to (0,0),\, x\neq y} \frac{x^2-xy}{\sqrt{x}-\sqrt{y}}}$ ;

(b) $\displaystyle{\lim_{(x,y)\to (2,0)}\frac{x^2-y^2-4x+4}{x^2+y^2-4x+4}}$

Solution.

(a)

$\begin{aligned} & \quad \lim_{(x,y)\to (0,0),\, x\neq y} \frac{x^2-xy}{\sqrt{x}-\sqrt{y}} \\ & = \lim_{(x,y)\to (0,0),\, x\neq y} \frac{x(x-y)}{\sqrt{x}-\sqrt{y}} \\ & = \lim_{(x,y)\to (0,0),\, x\neq y} \frac{x(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{x}-\sqrt{y}} \\ & = \lim_{(x,y)\to (0,0),\, x\neq y} x(\sqrt{x}+\sqrt{y}) \\ & = (0)\big( \sqrt{(0)} + \sqrt{(0)} \big) \\ & = 0 \end{aligned}$

(b)

Roughwork.

If we take limits along the path $(2,y)\to (2,0)$ ,

$\begin{aligned} & \quad \lim_{(x,y)\to (2,0)}\frac{x^2-y^2-4x+4}{x^2+y^2-4x+4} \\ & = \lim_{(x,y)\to (2,0)} \frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{y\to 0\textrm{ along }x=2}\frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{y\to 0\textrm{ along }x=2}\frac{\big((2)-2\big)^2-y^2}{\big((2)-2\big)^2+y^2} \\ & = \lim_{y\to 0\textrm{ along }x=2}\frac{-y^2}{y^2} \\ & = -1 \end{aligned}$

whereas if we take limits along the path $(x,0)\to (2,0)$ ,

$\begin{aligned} & \quad \lim_{(x,y)\to (2,0)} \frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{x\to 2\textrm{ along }y=0}\frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{x\to 2\textrm{ along }y=0}\frac{(x-2)^2-(0)^2}{(x-2)^2+(0)^2} \\ & = 1 \end{aligned}$

The limit fails to exist because the limiting values vary with the paths of taking the limit $\lim_{(x,y)\to (2,0)}$ .

April 18, 2021October 24, 2022

202104181519 Homework 1 (Q4)

i. Find the infinitesimal small vector $\mathbf{dr}$ in the cylindrical coordinate induced by an infinitesimal small changes of $\mathrm{d}\rho$ , $\mathrm{d}\theta$ , and $\mathrm{d}z$ in terms of $\rho$ , $\theta$ , $z$ , $\mathrm{d}\rho$ , $\mathrm{d}\theta$ , $\mathrm{d}z$ and the corresponding unit vector.

ii. $f(u_1, u_2, u_3)$ is defined in $\mathbf{r}=(u_1, u_2, u_3)$ coordinate. Its gradient is defined

$\displaystyle{\lim_{\Delta l_i\to 0}\sum_{i=1}^{3}\frac{\Delta f_i}{\Delta l_i}\hat{\mathbf{u}}_l}$

where $\Delta l_i$ and $\Delta f_i$ are respectively the changes in length and functional value induced purely by the infinitesimal change in $u_i$ . $\hat{\mathbf{u}}_l$ is the unit vector of $\mathbf{u}_i$ . Thus find the gradient of $f$ in cylindrical coordinate.

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2155 Methods of Physics II Homework Solutions.)

i.

$\mathbf{dr}=\mathrm{d}\rho\,\hat{\boldsymbol{\rho}}+\rho\,\mathrm{d}\theta\,\hat{\boldsymbol{\theta}}+\mathrm{d}z\,\hat{\mathbf{z}}$

Compare to the figure below.

ii.

$\begin{aligned} \nabla f & = \lim_{\Delta l_i\to 0}\sum_{i=1}^{3}\frac{\Delta f_i}{\Delta l_i}\hat{\mathbf{u}_i} \\ & = \lim_{\Delta\rho\to 0} \frac{\Delta f_\rho}{\Delta \rho}\,\hat{\boldsymbol{\rho}} + \lim_{\Delta\theta\to 0} \frac{\Delta f_\theta}{\rho\Delta\theta}\,\hat{\boldsymbol{\theta}} + \lim_{\Delta z\to 0}\frac{\Delta f_z}{\Delta z}\,\hat{\mathbf{z}} \\ & = \frac{\partial f}{\partial \rho}\,\hat{\boldsymbol{\rho}} + \frac{1}{\rho}\frac{\partial f}{\partial \theta}\,\hat{\boldsymbol{\theta}} + \frac{\partial f}{\partial z}\,\hat{\mathbf{z}} \end{aligned}$

April 16, 2021October 24, 2022

202104162114 Homework 1 (Q3)

Suppose $z=f(x,y)$ is differentiable. Find the expression

$\displaystyle{\bigg( \frac{\partial z}{\partial x} \bigg)^2 + \bigg( \frac{\partial z}{\partial y} \bigg)^2}$

in terms of polar coordinates $r$ and $\theta$ .

Solution.

Setup.

$\begin{aligned} & \begin{cases} \displaystyle{\frac{\partial z}{\partial r}} & = \displaystyle{\frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}} \\ \displaystyle{\frac{\partial z}{\partial \theta}} & = \displaystyle{\frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}} \\ \end{cases} \\ \Rightarrow & \begin{cases} \displaystyle{\frac{\partial z}{\partial r}} & = \displaystyle{\cos\theta\frac{\partial z}{\partial x} + \sin\theta\frac{\partial z}{\partial y}} \\ \displaystyle{\frac{\partial z}{\partial \theta}} & = \displaystyle{-r\sin\theta\frac{\partial z}{\partial x} + r\cos\theta\frac{\partial z}{\partial y}} \\ \end{cases} \\ \Rightarrow & \begin{bmatrix} \displaystyle{\frac{\partial z}{\partial r}} \\ \displaystyle{\frac{\partial z}{\partial \theta}}\end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{bmatrix} \begin{bmatrix} \displaystyle{\frac{\partial z}{\partial x}} \\ \displaystyle{\frac{\partial z}{\partial y}} \end{bmatrix} \\ \end{aligned}$

(to be continued)

Lemma.

_{Wikipedia on Cramer’s rule}

(continue)

$\begin{aligned} \frac{\partial z}{\partial x} & = \frac{\begin{vmatrix}\displaystyle{\frac{\partial z}{\partial r}} & \sin\theta \\ \displaystyle{\frac{\partial z}{\partial \theta}} & r\cos\theta \end{vmatrix}}{\begin{vmatrix}\cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{vmatrix}} \\ \\ & = \frac{r\cos\theta\displaystyle{\frac{\partial z}{\partial r}}-\sin\theta\displaystyle{\frac{\partial z}{\partial \theta}}}{r\cos^2\theta + r\sin^2\theta} \\\\ & = \cos\theta\displaystyle{\frac{\partial z}{\partial r}} - \frac{1}{r}\sin\theta\displaystyle{\frac{\partial z}{\partial \theta}} \\\\ \frac{\partial z}{\partial y} & = \frac{\begin{vmatrix}\cos\theta & \displaystyle{\frac{\partial z}{\partial r}} \\ -r\sin\theta & \displaystyle{\frac{\partial z}{\partial \theta}} \end{vmatrix}}{\begin{vmatrix}\cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{vmatrix}} \\\\ & = \frac{1}{r}\cos\theta\frac{\partial z}{\partial \theta} + \sin\theta\frac{\partial z}{\partial r} \end{aligned}$

Thus,

$\begin{aligned} & \quad\bigg( \frac{\partial z}{\partial x}\bigg)^2 + \bigg( \frac{\partial z}{\partial y}\bigg)^2 \\ \\ & = \bigg[ \cos\theta\frac{\partial z}{\partial r} - \frac{1}{r}\sin\theta\frac{\partial z}{\partial \theta} \bigg]^2 + \bigg[ \frac{1}{r}\cos\theta\frac{\partial z}{\partial \theta} + \sin\theta\frac{\partial z}{\partial r} \bigg]^2 \\ \\ & = \cos^2\theta \bigg( \frac{\partial z}{\partial r}\bigg)^2 - \frac{\sin 2\theta}{r}\bigg(\frac{\partial z}{\partial r}\bigg)\bigg(\frac{\partial z}{\partial \theta}\bigg) + \frac{1}{r^2}\sin^2\theta \bigg(\frac{\partial z}{\partial \theta}\bigg)^2 \\ & \qquad\qquad + \frac{1}{r^2}\cos^2\theta \bigg(\frac{\partial z}{\partial \theta}\bigg)^2 + \frac{\sin 2\theta}{r}\bigg(\frac{\partial z}{\partial r}\bigg)\bigg(\frac{\partial z}{\partial \theta}\bigg) + \sin^2\theta \bigg( \frac{\partial z}{\partial r}\bigg)^2 \\ \\ & = \bigg( \frac{\partial z}{\partial r} \bigg)^2 + \frac{1}{r^2} \bigg( \frac{\partial z}{\partial \theta} \bigg)^2 \end{aligned}$

April 16, 2021October 24, 2022

202104162147 Homework 1 (Q2)

The angle $a$ of a triangle $ABC$ is increasing at a rate of $3\,\mathrm{^\circ\, s^{-1}}$ , the side of $AB$ is increasing at a rate of $1\,\mathrm{cm\, s^{-1}}$ , and the side of $AC$ is decreasing at a rate of $2\,\mathrm{cm\, s^{-1}}$ . How fast is the side $BC$ changing when $a=30^\circ$ , $AB=10\,\mathrm{cm}$ , and $AC=24\,\mathrm{cm}$ ? Is the length of $BC$ increasing or decreasing?

Solution.

Draw a figure below:

Rephrase the problem.

Given that If , , and , then

By cosine law,

$z^2=x^2+y^2-2xy\cos a$ .

Taking ordinary derivatives w.r.t. time $t$ ,

$\displaystyle{2z\frac{\mathrm{d}z}{\mathrm{d}t} = 2x\frac{\mathrm{d}x}{\mathrm{d}t} + 2y\frac{\mathrm{d}y}{\mathrm{d}t} + 2xy\sin a\frac{\mathrm{d}a}{\mathrm{d}t} - 2x\cos a\frac{\mathrm{d}y}{\mathrm{d}t} - 2y\cos a\frac{\mathrm{d}x}{\mathrm{d}t}}$

$\begin{aligned} z & =\sqrt{x^2+y^2-2xy\cos a} \\ & = \sqrt{(10)^2+(24)^2-2(10)(24)\cos 30^\circ} \\ & = 16.1341\qquad (4\,\mathrm{d.p.}) \end{aligned}$

Plugging in the value of each,

$\displaystyle{2(\cdot\cdot )\frac{\mathrm{d}z}{\mathrm{d}t} = 2(\cdot\cdot )\big(\cdot\cdot \big)+2(\cdot\cdot )\big(\cdot\cdot \big)+2(\cdot\cdot )(\cdot\cdot )\sin (\cdot\cdot )\big(\cdot\cdot \big) - 2(\cdot\cdot )\cos (\cdot\cdot )\big(\cdot\cdot \big) - 2(\cdot\cdot )\cos (\cdot\cdot )\big(\cdot\cdot \big)}$

you will know what $\displaystyle{\frac{\mathrm{d}z}{\mathrm{d}t}}$ is.

But now, I intend to treat it with partial derivatives.

Let $f(x,y,a) = x^2+y^2-2xy\cos a = z^2$ .

$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}t} & = \frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{\partial f}{\partial a}\frac{\mathrm{d}a}{\mathrm{d}t} \\ & = (2x-2y\cos a)\frac{\mathrm{d}x}{\mathrm{d}t} + (2y-2x\cos a)\frac{\mathrm{d}y}{\mathrm{d}t} + 2xy\sin a\frac{\mathrm{d}a}{\mathrm{d}t} \\ \end{aligned}$

After $\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}t}}$ is sought, recognise that

$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}t} & =2z\frac{\mathrm{d}z}{\mathrm{d}t} \\ \frac{\mathrm{d}z}{\mathrm{d}t} & = \bigg(\frac{1}{2z}\bigg)\frac{\mathrm{d}f}{\mathrm{d}t} \end{aligned}$

you could have it also.

(to be continued)

April 16, 2021October 24, 2022

202104160754 Homework 1 (Q3)

A ship $A$ , which can sail at a constant speed $60\,\mathrm{km/hr}$ to meet a second ship $B$ which is $100\,\mathrm{km}$ away in the direction of $\mathrm{S60^\circ W}$ and is sailing due east at constant speed $30\,\mathrm{km/hr}$ . Find the sailing direction of $A$ and the time required to meet $B$ .

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

Draw a diagram as follows:

Setup.

By the law of sines,

$\begin{aligned} \frac{V_A}{\sin 30^\circ} & = \frac{V_B}{\sin\theta} \\ \frac{60}{\sin 30^\circ} & = \frac{30}{\sin\theta} \\ \sin\theta & = 0.25 \\ \theta & = 14.5^\circ \end{aligned}$

Direction of $\mathbf{v}_A$ : $\mathrm{S45.5^\circ W}$

$\because 180^\circ -30^\circ -14.5^\circ - 90^\circ = 45.5^\circ$

Calculating $v_{AB}$ :

$\begin{aligned} |\mathbf{v}_{AB}| & = |\mathbf{v}_{A}|\cos\theta + |\mathbf{v}_{B}|\cos 30^\circ \\ & = 60 \cos 14.5^\circ + 30\cos 30^\circ \\ & = 84.1\,\mathrm{km/hr} \end{aligned}$

The time needed to meet ship $B$ is

$\begin{aligned} t & = \frac{100\,\mathrm{km}}{|\mathbf{v}_{AB}|} \\ & = \frac{100\,\mathrm{km}}{84.1\,\mathrm{km/hr}} \\ & = 1.19\,\mathrm{hr} \end{aligned}$

April 16, 2021October 24, 2022

202104160620 Homework 1 (Q4)

A particle is projected from a point $O$ on the horizontal floor. The range of the projectile is $R$ and the maximum height that the particle can reach is $h$ . Show that the equation of trajectory of the particle is

$\displaystyle{\frac{y}{h}=\frac{4x}{R}\bigg( 1-\frac{x}{R}\bigg)}$ .

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

The trajectory of projectile motion must be a parabola, which can be expressed in the form of a quadratic equation:

$y=ax^2+bx+c$ ;

And since the particle passes through the points $(0,0)$ and $(R,0)$ , the equation of trajectory can be expressed in the form:

$y=A(x-0)(x-R)$ .

When the particle has traveled a horizontal distance $x=\displaystyle{\frac{R}{2}}$ , it reaches the maximum height $y=h$ .

Inserting the point $(\frac{R}{2},h)$ into the trajectory equation, we solve for the unknown $A$ :

$\begin{aligned} h & = A\bigg(\frac{R}{2}-0\bigg)\bigg(\frac{R}{2}-R\bigg) \\ h & = -\frac{AR^2}{4} \\ \Rightarrow \qquad A & = -\frac{4h}{R^2} \end{aligned}$

Thus,

$y = -\displaystyle{\frac{4h}{R^2}}x(x-R)$ ,

or,

$\boxed{\frac{y}{h} = \frac{4x}{R}\bigg( 1-\frac{x}{R}\bigg)}$