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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 $Vy$.


(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??}

ball from platform

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You have too many constraints ($h$,$d$,$H\%$ and $D\%$) and too few controls ($v$ and $A$) – Henry Jul 3 '12 at 8:14
Note that $D\leq 50\%$, since presumably $h\geq 0$. – Alex Becker Jul 3 '12 at 8:22
It's impossible to have $D\% = 50$ unless $h = 0$. At a horizontal displacement of $2D\%$, the projectile must return to its original height $h$. – Rahul Jul 3 '12 at 8:35
However, I think $D\%$ can be made arbitrarily close to $0$ or $50$ by making $v\sin\alpha$ very small or very large. – Rahul Jul 3 '12 at 8:43
up vote 3 down vote accepted

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 $$v_x = \frac{dg}{v_y + \sqrt{v_y^2+2gh}}.$$

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@JoeBlow, the explanation can't have been that superb if it didn't get my point across! I've edited the answer. In general, there is no solution that satisfies all your requirements. – Rahul Jul 3 '12 at 17:54
@Joe: Yes, exactly. Either (1) or (2) will give you the value of $v_y$, but in general their answers will be different, which is why you can't fix both $H$ and $D$ independently. You still need (3) to get the value of $v_x$ after you have $v_y$. – Rahul Jul 3 '12 at 19:22
Joe Blow tried to edit this into the answer, but I think it is more appropriate as a comment so I move it here below. – Ragib Zaman Jul 4 '12 at 12:56
Note for future readers: to solve 'ProblemAA', you must use Rahul's equations 1 & 3. To solve 'ProblemBB', you must use Rahul's equations 2 & 3. Kindly, Joe. – Ragib Zaman Jul 4 '12 at 12:56

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 $H\%$.

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@Joe Blow you are welcome :). well Rahul didn't derive $v_x$ I think.And for your Note I have written my Note that $D\% >50\%$ is possible. – Saurabh Jul 3 '12 at 15:21
I don't think you're satisfying the condition that the ball must land at a distance $d$. (This is not the same as the condition that the ball achieves its maximum height at $x$-coordinate $d\frac D{100}$.) – Rahul Jul 3 '12 at 17:57
Now your solution is inconsistent. You're trying to choose all three of $\tan\alpha$, $v_y$, and $v_x$ independently, which doesn't make sense because $\tan\alpha$ has to be $v_y/v_x$. – Rahul Jul 3 '12 at 18:55
@RahulNarain I think it is the same.It is just that now $D$ is not independent that's why "result point" wildly varies. Let $t_1$ and $t_2$ denotes time taken in 2 halves of the journey. $$t_1=\frac{v_y}{g} \text{ and } t_2= \sqrt{\frac{2h}{g}+\frac{{v_y}^2}{g^2}}$$ then $$\frac{d-\frac{dD}{100}}{\frac{dD}{100}}=\frac{t_2}{t_1}$$ from which we get $$\frac{100}{D}=\frac{v_y+\sqrt{{v_y}^2+2gh}}{v_y}$$ And then substituting in $\frac{Ddg}{100v_y}$ we get the same expression for $v_x$ That means if you had choose $h,d$ and $H$ then $v_y$ get fixed and then $D$ cannot be choosen independently – Saurabh Jul 3 '12 at 19:00
But still I had adopted @RahulNarain idea because from my previous expression($v_x=\frac{Ddg}{100v_y}$) it seems that $d$ and $D$ can be arbitarily choosen which is not the case when $H$and $h$ is fixed. – Saurabh Jul 3 '12 at 19:04

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