TITLE: What is minimum speed needed to jump over sphere object that has radius R and at distance d?
QUESTION [18 upvotes]: (I am not expert in English. I will write as well as I can.)
To understand this question easier, lets see this picture.

From this picture, what is minimum initial speed that this grasshopper need to jump over this log?
The grasshopper movement path can touch the log but can't cross through inside of the log.
d and R can be any positive real number which d>R. g is a gravitational acceleration(approximately 9.80665 $m/s^2$).
This is a mathematical-physics question but mainly in maths.
I can do physics part but have problem in maths part.
Physics Part : Let red ball is grasshopper and at origin point and y is height


The relation between x and y for projectile motion is $y(x) = xtan\theta - \frac{gx^2}{2u^2cos^2\theta}$ , $0 < \theta < \frac{\pi}{2}$
Upper curve of sphere can de describe in function : $y_s(x) = R + \sqrt{R^2-(x-d)^2}$ , for $ d-R \leq x \leq d+R$
Condition of the sphere is $y(x) \geq y_s(x)$ , for $ d-R \leq x \leq d+R$ or
$xtan\theta - \frac{gx^2}{2u^2cos^2\theta} \geq R + \sqrt{R^2-(x-d)^2}$ , for $ d-R \leq x \leq d+R$
At this point, I don't how to find $u_{min}(d,R)$ from this.
(If you give value of d and h (for example, d = 2m and R = 1m), it is possible to find $\theta$ that minimize u.)
I know only that y(x) (parabola curve) for $u_{min}(d,R)$ look like.
Case : d = cR (c is a constant. There is a ratio that make y(x) have maximun point at the top of the sphere.)

Case : d > cR

Case : d < cR

Please help me.

REPLY [2 votes]: For $R<d\le 2R$ and symmetrical contact (two contact points, on either side of circle)
$\hspace{1cm}$
Consider the case where the parabola (projectile trajectory) touches the circle (cross section of sphere) at two points, $A, B$.  Let these points subtend an angle $\theta$ to the vertical at the centre of the circle.
Take the contact point between the circle and ground as the origin. 
Let $V(0,h)$ be the point of intersection of the tangents (to both trajectory and circle) at $A,B$. By simple trigonometry, it can be easily worked out that $h=R\left(1+\frac 1{\cos\theta}\right)$
Hence the three points are
$$A\big(-R\sin\theta, R(1+\cos\theta)\big); \qquad V\big(0,R\left(1+\tfrac 1{\cos\theta}\right)\big); \qquad 
B\big(R\sin\theta, R(1+\cos\theta)\big)$$
Hence the parametric form of the parabola can be stated as a quadratic Bezier curve as follows:
$$\begin{align}
(x,y)
&=\bigg(t^2(-R\sin\theta+2t(1-t)\cdot 0+(1-t)^2(R\sin\theta)\; ,\\
&\qquad t^2 R(1+\cos\theta)+2t(1-t)R(1+\tfrac 1{\cos\theta})+(1-t)^2R(1+\cos\theta) \bigg)\\
&=\bigg(\ (1-2t)R\sin\theta\; ,\;(1-2t+2t^2)R(1+\cos\theta)+2t(1-t)R(1+\tfrac 1{\cos\theta})\ \bigg)\end{align}$$
Eliminating $t$ gives the Cartesian form of the trajectory
$$\boxed{y=\frac {R^2(1+\cos\theta)^2-x^2}{2R\cos\theta}}\\
\text{which is equivalent to }\\
\boxed{\frac {x^2}{R^2(1+\cos\theta)^2}+\frac y{\left(\frac {R^2(1+\cos\theta)^2}{2R\cos\theta}\right)}=1}$$
As the launch point is given as $(-d,0)$, this gives
$$d=R(1+\cos\theta)\Longrightarrow \cos\theta=\tfrac dR-1$$
Hence the equation of the trajectory becomes
$$\boxed{y=\frac {d^2-x^2}{2(d-R)}}\\
\scriptsize(R<d\le 2R)\\
\text{which is equivalent to}\\
\boxed{\frac {x^2}{d^2}+\frac y{\left(\frac{d^2}{2(d-R)}\right)}=1}\\
\frac {dy}{dx}=-\frac x{d-R}=\frac d{d-R}\text{ at } x=-d$$
Let $v, \alpha$ be the launch velocity and launch angle respectively.
Hence $$\boxed{\tan\alpha =\frac d{d-R}}$$
Equating coefficients of $x^2$ with the standard trajectory equation $y=x\tan\alpha-\frac {gx^2}{2v^2}(1+\tan^2\alpha)$ gives
$$\begin{align}
\frac 1{2(d-R)}
&=\frac g{2v^2}(1+\tan^2\alpha)\\
&=\frac g{2v^2}\big(1+\left(\tfrac d{d-R}\right)^2\big)\\
\color{red}{v^2}&\color{red}{=gd\left(\frac d{d-R}+\frac {d-R}d\right)}
\qquad \scriptsize (R<d\le 2R)
\end{align}$$
Another method:
Parabola:
$$\frac {x^2}{d^2}+\frac yh=1\tag{1}$$
Circle:
$$x^2+(y-R)^2=R^2\tag{2}$$
$d^2\times (1)-(2)$:
$$\begin{align}
(y-R)^2-\frac {d^2}hy&=R^2-d^2\\
y^2-(2R+\tfrac{d^2}h)y+R^2=R^2-d^2\\
y^2-\left(\frac {2Rh+d^2}h\right)y+d^2&=0\\
\text{For tangency,  }
\left(\frac {2Rh+d^2}h\right)^2&=4d^2\\
\frac {2Rh+d^2}h&=\pm 2d\\
h&=\frac {d^2}{2(d-R)}\end{align}$$
Hence equation for parabola is 
$$\boxed{y=\frac {d^2-x^2}{2(d-R)}}$$
and the rest follows per above. 

For $d\ge 2R$ and symmetrical contact (one contact point, at top of circle)

For this case where $d\ge 2R$, the launch point is sufficient far enough to just touch the top of the circle without hitting any other part. Hence this is the same as a projectile with its vertex of its trajectory at the top of the circle. 
Using standard formulas for maximum height and range of a projectile, we have
$$\begin{align}
2R&=\frac {v^2}{2g}\sin^2\alpha\qquad(1)\\
d&=\frac {v^2}{2g}\sin 2\alpha\qquad(2)\\
(1)/(2):\hspace{1cm}\frac{2R}d&=\frac {\tan\alpha}2\\\end{align}$$
$$\boxed{\tan\alpha=\frac {4R}d}\\$$
$$\begin{align}\text{From }(2):\hspace{1cm}\\
d&=\frac {v^2}{2g}\cdot \frac {2\tan\alpha}{1+\tan^2\alpha}\\
\color{red}{v^2}&\color{red}{=gd\left(\frac {d}{4R}+\frac{4R}d\right)}
\qquad \scriptsize (d\ge 2R)
\end{align}$$
The equation of the parabola is
$$\boxed{\frac{x^2}{d^2}+\frac y{2R}=1}\\
\text{which is equivalent to}\\
\boxed{y=2R\left(1-\frac {x^2}{d^2}\right)}$$

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