How to find the max height a projectile can reach on a hillside

[Justinjah91]

Homework Statement

You see an enemy archer and decide to position yourself on the side of a hill with a 60 degree grade. The enemy archer is 300 meters from the base of the hill. If the enemy's bow fires arrows at 60 m/s, how high (measured vertically) do you need to be on the hill to be out of range of the enemy archer?

Relevant Equations

Kinematics: $x(t)=x_i+v_it+\frac 1 2 at^2$

Here's a fully typed version of the problem with a diagramMy attempt:
Given the angle of the hill, I know that the horizontal displacement of the arrow and my vertical height on the hill are related by

$Δx=d+\frac h {Tan(60)}$ ...(1)​

​

where d is the distance of the enemy from the base of the hill (300 m). I also know that there is no acceleration in the horizontal direction, and the time of flight of the arrow is the same for the horizontal and vertical directions, leading to the following:

$Δx=v_{xi}t~~$ ⇒ $~~t=\frac {Δx} {v_xi}$​

​

$Δy=v_{yi}t-\frac 1 2 gt^2~~$ ⇒ $~~Δy=v_{yi}\left( \frac {Δx} {v_xi} \right) - \frac 1 2 g \left( \frac {Δx} {v_{xi}} \right)^2$ ...(2)​

​

If I now substitute eqn (1) into (2), and let $Δy=h$, I get:

$h=v_{yi} \left( \frac {d+ \frac h {Tan(60)}} {v_{xi}}\right)-\frac 1 2 g\left( \frac {d+ \frac h {Tan(60)}} {v_{xi}}\right)^2$​

​

Lastly, I know that the components of my initial velocity are related to the launch angle by:

$v_{yi}=v_isinθ$​

$v_{xi}=v_icosθ$​

Now, to find the maximum height, I need to maximize "h" as a function of theta by taking dh/dθ=0. The problem is that I'm having trouble getting "h" as a function of "θ". I'm getting stuck in the algebra somewhere. Anybody have a handy way to simplify this a little further? (I drew it as a cannon on the whiteboard, but same idea)

I feel like there must be a trig substitution or something that I could have employed much earlier in this process to make my life easier.

 

[TSny]

When $h$ reaches its maximum, does $\Delta x$ also attain its maximum? If so, what does that tell you about $\frac {dh}{d\theta}$ and $\frac {d (\Delta x)}{d \theta}$ when these derivatives are evaluated at the value of $\theta$ that corresponds to maximum $h$?

 

[Justinjah91]

Yes I suppose it does. That would mean that $\frac {dh} {dθ} = \frac {d(Δx)} {dθ} =0$.

I've gone ahead and given that a try, but the displacement I get is dependent on that optimum launch angle, and I still don't know that. I'm fairly certain that I've done the derivatives correctly, but maybe I'm missing something?

 

[TSny]

OK. We ask that you try to avoid just posting pictures of your work. It makes it difficult for helpers to quote specific parts of your work. General guidelines are given here:
https://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686781/
I agree with your result for the relation between $\Delta x$ and $\tan \theta$.

You can use it to replace $\tan \theta$ in terms of $\Delta x$ in the first equation in the picture of your last post. Note $\frac{1}{\cos ^2 \theta} = 1+ \tan ^2 \theta$.

 

[Justinjah91]

Ah yeah that makes sense. Sorry about that!

Unfortunately, plugging that in makes it blow up into a crazy 4 order polynomial that I have no clue how to solve. I'm probably going to just use a numerical solver rather than trying for the analytical solution. I really would have expected it to be a nicer outcome than this.

Thank you so much for the help though!

 

[TSny]

You should find quite a bit of simplification after substituting for $\tan \theta$. You should get a quadratic equation for $\Delta x$ (or for $h$, if you prefer).

 

[hurreechunder]

The max range for 45 deg launch is 366 metres. The halfway point is thus 183 m. Therefore, the projectile will hit the hill on its way down rather than up. Up to 45 deg, all projectile paths intersect the 45 deg trajectory behind its point of max height (or its halfway point along the x-axis). Therefore, the max point on the hill corresponds to the path of the 45 deg projectile.

 

[hurreechunder]

It's actually trickier if the distance to the base is <183 m. In that case, the max height does not correspond to the 45 deg projectile path and a different approach has to be used.

 

[TSny]

Hello, hurreechunder.

Up to 45 deg, all projectile paths intersect the 45 deg trajectory behind its point of max height (or its halfway point along the x-axis).

A projectile fired at less than 45^o does not intersect the trajectory fired at 45^o (except at the launch point, of course).

Therefore, the max point on the hill corresponds to the path of the 45 deg projectile.

Did you consider trajectories with launch angles greater than 45^o?

 

[jbriggs444]

Up to 45 deg, all projectile paths intersect the 45 deg trajectory behind its point of max height (or its halfway point along the x-axis).

Between 0 and 45 degrees, the second intersection point would be somewhere below the shooter's level. If the ground is level, that puts the second intersection point underground somewhere beyond max range.

Between 45 and 90 degrees, the second intersection point could be either on the rising portion of the 45 degree trajectory or on the falling side.

The first intersection point is at the shooter's position, obviously.

 

[hurreechunder]

Hello, hurreechunder.
A projectile fired at less than 45^o does not intersect the trajectory fired at 45^o (except at the launch point, of course).

Did you consider trajectories with launch angles greater than 45^o?

Sorry, I meant from 90 deg to 45 deg, not from 0 deg to 45 deg. Hope that clarifies

 

[hurreechunder]

Between 0 and 45 degrees, the second intersection point would be somewhere below the shooter's level. If the ground is level, that puts the second intersection point underground somewhere beyond max range.

Between 45 and 90 degrees, the second intersection point could be either on the rising portion of the 45 degree trajectory or on the falling side.

The first intersection point is at the shooter's position, obviously.

You’re right, it’s possible for the intersection point to be on the falling side as well. I didn’t see that.
The “envelope of safety” formula should provide the answer in that case.

 

[BvU]

The “envelope of safety” formula

Didn't see that formula come by
Can you show us the quadratic equation you found ?