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})##.