Projectile motion - determining initial velocity and angle

[marksyncm]

Homework Statement

An archer launches an arrow from coordinates $(0, 0)$ at an angle $\alpha$ and with an initial velocity $v_0$. There's a target located ahead of the archer and the center of that target is at coordinates $(d, h)$. At what $v_0$ and at what angle $\alpha$ does the archer need to launch his arrow to hit the center of the target perpendicular to its surface?

Homework Equations

Kinematic equations

The Attempt at a Solution

We have that:

$v_x = v_0\cos(\alpha)$
$v_y = v_0\sin(\alpha) - gt$

For the arrow to hit the center of the target perpendicular to its surface, it must have a vertical velocity of zero when it reaches the target. So the following conditions must be met:

1) By the time the arrow travels a distance $d$ in the $x$ direction, its vertical velocity component should be equal to zero.
2) By that same time above, the arrow should be located at height = $h$.

So we have:

$d = v_0\cos(\alpha)t_i \rightarrow t_i = \frac{d}{v_0\cos(\alpha)}$ ........ (1)
$0 = v_0\sin(\alpha) - gt_i \rightarrow t_i = \frac{v_0\sin(\alpha)}{g}$ ...... (2)
$\frac{d}{v_0\cos(\alpha)} = \frac{v_0\sin(\alpha)}{g} \rightarrow v_0^2\sin(\alpha)\cos(\alpha) = dg$ .... (3)

Where $t_i$ is the time of impact.

I'm at a loss as to how to proceed from here. I'm assuming I need to use the displacement equation in the vertical direction: $h = v_0\sin(\alpha)t - \frac{g}{2}t^2$, but I am not sure what to put into this equation. Do I put in the $t_i$ from equation (1) or from equation (2)? Can I use both values of $t_i$ simultaneously (insert one under $t$ and another under $t^2$) since they are supposed to be the same thing? I tried doing this and I get an algebraic expression that I'm unable to process. For example, here's what happens when I use $t_i$ from equation (1):

$$h=\frac{v_0\sin(\alpha)d}{v_0\cos(\alpha)} - \frac{gd^2}{2v_0^2\cos^2(\alpha)} \rightarrow h = \frac{sin(\alpha)}{\cos(\alpha)} - \frac{gd^2}{2v_0^2cos^2(\alpha)}$$

But this equation doesn't account for the fact that at time $t_i$, the vertical velocity needs to be zero. How do I proceed?

 

[Orodruin]

Your image suggests that the target is a sphere, but your solution and the formulation suggests that it is a vertical surface. I am going to assume the latter.

The value of $t_i$ must be the same in both cases. If not, your conditions would not be satisfied. However, I would suggest a different approach than using $v_0$ and $\alpha$ from the beginning. Instead, I would use the initial $x$ and $y$ components of the velocity (call them $v_{0x}$ and $v_{0y}$ for example). Either way, you should have a sufficient number of conditions to solve for all of your variables. In principle, you have three conditions and three unknowns:

1. The vertical velocity needs to be zero at $t_i$.
2. The vertical distance needs to be $h$ at $t_i$.
3. The horizontal distance needs to be $d$ at $t_i$.

This is a solvable system of equations.

 

[marksyncm]

Your image suggests that the target is a sphere, but your solution and the formulation suggests that it is a vertical surface. I am going to assume the latter.

Sorry for the confusion. Your assumption is correct.

Thanks, I'll take another stab at the problem using your approach and will see how it goes.

 

[marksyncm]

Here's what I did:

Time when vertical velocity = 0:

$$0 = v_{0y} - gt_i \rightarrow t_i = \frac{v_{0y}}{g}$$

Vertical distance is $h$ at time = $t_i$:

$$h = v_{0y}t - \frac{g}{2}t^2 \rightarrow H=v_{0y}\frac{v_{0y}}{g} - \frac{g}{2}\frac{v_{0y}^2}{g^2} = \frac{v_{0y}^2}{g}-\frac{v_{0y}^2}{2g} = \frac{v_{0y}^2}{2g}=h$$

Horizontal distance is $d$ at time = $t_i$:

$$d=v_{0x}t \rightarrow d=v_{0x}\frac{v_{0y}}{g} \rightarrow d=\frac{v_{0x}v_{0y}}{g}$$

So we have two equations:

$h = \frac{v_{0}^2 sin^2(\alpha)}{2g}$ ...... (1)
$d=\frac{v_{0}^2 \sin \alpha cos \alpha}{g}$ ...... (2)

From equation (2), we get that $v_{0}^2 = \frac{dg}{\sin \alpha \cos \alpha}$. Substituting this into equation (2):

$$h = \frac{dg \sin^2(\alpha)}{2g\sin(\alpha)\cos(\alpha)} = \frac{d\sin(\alpha)}{2\cos(\alpha)} = \frac{d}{2} \tan(\alpha) = h \rightarrow \frac{2h}{d} = \tan(\alpha) \iff \alpha = tan^{-1}(\frac{2h}{d})$$

Is this correct?

However, I'm not sure how to get the value for $v_0$ from equations (1) and (2). It seems that whatever substitution I make ends up with a $v_0^2$ in the numerator canceling out with another $v_0$ in the denominator. Would appreciate a tip.

 

[Orodruin]

Well, $\alpha$ is now known so you can just substitute it into either (1) or (2) and solve for $v_0$.

Edit: ... or you could just solve for $v_{0x}$ and $v_{0y}$ and apply Pythagoras' theorem.

 

[marksyncm]

Well, $\alpha$ is now known so you can just substitute it into either (1) or (2) and solve for $v_0$.

Edit: ... or you could just solve for $v_{0x}$ and $v_{0y}$ and apply Pythagoras' theorem.

Thank you!

Is it "unusual" that I cannot find the value of $v_0$ from just equations (1) and (2) (without substituting in $\alpha$)? Or am I just not seeing a way to do it?

 

[Orodruin]

No, it is not unusual in any way because $v_0$ has contributions from both the vertical and the horizontal velocity components.

 

[ehild]

Thank you!

Is it "unusual" that I cannot find the value of $v_0$ from just equations (1) and (2) (without substituting in $\alpha$)? Or am I just not seeing a way to do it?

You can use the trigonometric identity $\sin^2(α)=\frac{\tan^2(α)}{1+\tan^2(α)}$ in equation (1).
Or you can apply the double-angle formulas sin²(α)=(1-cos(2α))/2, sin(α)cos(α)=sin(2α)/2,
isolate sin(2α) and cos(2α), square and add the squares, resulting 1. You get an equation for v₀².

 

[gneill]

I think I would have begun with conservation of energy to get the vertical velocity component:

$\frac{1}{2}v_y^2 = gh$ so that $v_y = \sqrt{2 g h}$

Then, since the time to rise to the maximum height is the same as for falling from that height,

$\frac{1}{2} g t^2 = h$ so that $t = \sqrt{2 \frac{h}{g}}$

$v_x$ then follows from the time and horizontal distance. The angle follows that from $\arctan(\frac{v_y}{v_x})$.