Ball from platform with specific vertex

Question

On earth in a vacuum. You throw a platon from a platform height $h$ and want it to land at point $d$ distant. Note, h is absolutely fixed and d is absolutely fixed. It "must land" at point d, no matter what. You throw it with velocity expressed using $Vx$ and $Vy$.

Now,

(Problem AA)

you want the vertex HEIGHT to be at a percentage $H$ >100 of $h$ (say $H=120 \% $).

-- ~~ ~~ OR ~~ ~~ --

(Situation BB)

you want the vertex DISTANCE to be at a percentage $D$ <50 of $d$ (say $D=25 \% $).

NOTE: the two gentlemen below have generously explained that you CANNOT choose BOTH H and D. Thank you very much for this insight and proof!

So! For each of ProblemAA and ProblemBB, how to calculate $Vx$ and $Vy$ ?

If this is possible - thank you!

{Aside: I assume there's only one $Vx$ / $Vy$ solution for a given value in either ProblmeAA or ProblemBB - but could there be two, or more??}

Answer 1

Instead of initial speed $v$ and angle $\alpha$, it's easier to work in Cartesian coordinates, with horizontal and vertical components of velocity $v_x$ and $v_y$.

The height of the vertex is $h + v_y^2/2g$, where $g$ is the acceleration due to gravity. To have a height of $h\cdot\frac H{100}$, you need $$v_y = \sqrt{2gh\left(\frac H{100}-1\right)}.$$

As the horizontal velocity is constant, the ratio of distance in the horizontal direction is the same as the ratio of the corresponding times. The vertical displacement of the projectile is $h + v_yt - \frac12gt^2$. The time to reach the vertex is $v_y/g$, while the total time to drop to height $0$ is $\big(v_y + \sqrt{v_y^2+2gh}\big)/g$. You want their ratio to be $\frac D{100}$, which implies that $$v_y = \sqrt{2gh\left(\frac{(D/100)^2}{1-2(D/100)}\right)}.$$

If the initial $h$ is fixed, you can't choose both $H$ and $D$ independently; there will be no solution that has both the $H$ you want and the $D$ you want, unless by coincidence they both give the same value of $v_y$.

The horizontal velocity of the projectile doesn't come into this at all. If you pick one of the equations, say you specify $H$ and decide you don't care about $D$, then after you have the solution for $v_y$ you can always choose $v_x$ to make the projectile land at distance $d$. In fact, since the time of flight is $\big(v_y + \sqrt{v_y^2+2gh}\big)/g$, and the projectile must travel a distance $d$ in that time, you can just set $$v_x = \frac{dg}{v_y + \sqrt{v_y^2+2gh}}.$$

Answer 2

Let $t$ be the time taken by the ball to reach to the vertex.
So in vertical direction$$v \sin \alpha -gt=0$$ $$t=\frac{v \sin \alpha}{g}$$ In horizontal direction$$\begin{align*}\frac{Dd}{100}&=v \cos \alpha.t\\ &=\frac{v^2 \cos \alpha \sin \alpha}{g}\quad \quad \cdots(1)\end{align*}$$ Again in vertical direction,applying $v^2=u^2+2as$ $$0=(v \sin \alpha)^2-2g ( \frac{Hh}{100}-h )$$ $$(v \sin \alpha)^2=2g ( \frac{Hh}{100}-h )\quad \quad \cdots(2)$$ Dividing $(2)$ and $(1)$ we get $$\tan \alpha = \frac{2h(H-100)}{Dd}$$ Similarly we can get $v$ also. $$v_y = \sqrt{2gh ( \frac{H}{100}-1)} \quad \cdots \cdots \text{from}(2)$$ and $$-v_yT+\frac{1}{2}gT^2=h$$ $$T=\frac{v_y+\sqrt{{v_y}^2+2gh}}{g}$$ where T= Time of flight
then $$v_x = \frac{dg}{v_y + \sqrt{v_y^2+2gh}}$$ Also it can be easily proved that if $D=50\%$ then $h=0$. So, $0\%<D\%\leq50\%$.

Note:If $h<0$(i.e.below the ground) is allowed then $D\%\geq50\%$ is possible but then we some what have to change the notion of $H\%$.