The distance between a boy and the shelf is R. He wants to throw a ball of mass m with an initial speed v such that it hits the top of the shelf of a height h.
1) Show that a ball thrown at the angle to the horizontal, then at any later time it has a horizontal displacement x and vertical displacement y (from the origin) satisfies:
$$y=x\tan\theta-\dfrac{gx^2}{2v^2\cos^2\theta}$$
Gravity is the only force in this example.
2) Show that the min speed that the ball needs to be thrown at is given by:
$$v^2=g\left(h+\sqrt{R^2+h^2}\right)$$
Thanks!
You can resolve the velocity into component forms $v\cos\theta$ and $v\sin\theta$. Since the horizontal is constant (assuming gravity is the only force acting), the horizontal displacement $x$ from the origin after time $t$ is given by $ x = v\cos\theta \cdot t$.
The vertical displacement $y$ is given by $$y = v\sin\theta \cdot t - \frac{gt^2}{2}$$
Substituting $t = \frac{x}{v\cos\theta}$ gives you (1)
Edit: sorry for the late completion as I didn't have an opportunity.
With the second part, you want to rearrange for $v^2$ and substitute $y$ and $x$ for $h$ and $R$ respectively.
Then we have $$\begin{align} v^2 & = \frac{gR^2}{2R\sin\theta\cos\theta - 2h\cos^2\theta} \\ & = \frac{gR^2}{R\sin2\theta - h(\cos2\theta+1)} \end{align}$$.
Thus the $v^2$ is a function of $\theta$. We can minimize it by finding the maximum value the denominator $d$ holds, or, in other words, finding the value of $\theta$ in which the derivative is zero (you should also check the sign of the second derivative or the signs on either side of the value at $\theta$). Then substitute this $\theta$ into the original equation.
Applying this we see that $d'(\theta) = 2R\cos2\theta + 2h\sin2\theta = 0.$ Thus $$\tan2\theta = -\frac{R}{h},$$ $$\sin2\theta = \frac{R}{\sqrt{R^2 +h^2}}, $$ and $$ \cos2\theta = \frac{-h}{\sqrt{R^2 +h^2}}.$$ Adding this to the equation for $v^2$ gives us
$$ v^2 = \frac{gR^2}{R\sin2\theta - h(\cos2\theta+1)} $$ $$ =\frac{gR^2}{\cfrac{R^2}{\sqrt{R^2 +h^2}} +\cfrac{h^2}{\sqrt{R^2 +h^2}} - h} $$ $$= \frac{gR^2}{\sqrt{R^2 + h^2}- h}$$ $$= \frac{(gR^2)(\sqrt{R^2 + h^2}+ h)}{(\sqrt{R^2 + h^2} - h)(\sqrt{R^2 + h^2}+h)}$$ $$ = g(\sqrt{R^2 + h^2}+ h).$$