Question

In a cherry-pit spitting competition, Ben stands at the edge of a hill that slopes downward at a 45 degree angle. He launches the pit with a velocity of 20 ft/sec, and his mouth is 6 feet above the ground. What angle Θ would maximize the distance that the pit is launched? What distance is reached? How long is the cherry pit in the air?
(diagram coming...)

Answer 1

Diagram is attached.

Answer 2

@mathslover @heena Can you help me?

Answer 3

We know a 45 degree launch angle maximizes range on a flat surface. I wonder if the same holds here. Let's check shall we?

First, set up our equations of motion of the pit. We will concern ourselves with the ground geometry in a bit.

\[\dot x_0 = v_0 \cos(\theta)\]\[\dot y_0 = v_0 \sin(\theta)\]where \(\dot x~ and ~ \dot y\) are the velocities in the x and y direction. x being horizontal and y being vertical.

I'll assume you have a grasp of the concepts of projectile motion and just present you with these equations\[y(t) = y_0 + \dot y_0 t - {1 \over 2} g t ^2\]\[x = R = \dot x_0 t\]

Now, we need to modify our equation for vertical displacement to account for the changing elevation of the hill. Let's realize that a 45 degree angle has a slope of 1. That is for every unit we move in y, we move one in x. \[y_{hill} = x\]\[y(t) = y_0 + y_{hill}(t) + \dot y_0 t - {1 \over 2} g t^2\]This is convenient because we are adding the downward displacement of the hill to the vertical displacement of the pit. This allows us to use the same methodology as used when proved that 45 is the maximum launch angle.

First, we know that when the pit hits the hill, \(y(t) = 0\) Therefore,\[0 = y_0 + y_{hill}(t) + \dot y_0 t - {1 \over 2} g t^2\]Let's substitute \(\dot x_0 t = y_{hill}\)\[0 = y_0 + (\dot x_0 + \dot y_0) t - {1 \over 2} g t^2\]Let's solve this for t, taking only the positive solution\[t = {-(\dot x_0 + \dot y_0) + \sqrt{(\dot x_0 + \dot y_0)^2 - 4(-0.5 g)(y_0)} \over g}\]Let's substitute in our expressions for \(\dot x ~ and~ \dot y\)\[t = {-(v_0(\sin(\theta) + \cos(\theta)) + \sqrt{(v_0(\sin(\theta) + \cos(\theta))^2 + 2gy_0} \over g}\]Let \(a=v_0(\cos(\theta) + \sin(\theta))\)\[t = {-a + \sqrt{v_0^2 (\sin(2\theta) + 1) + 2gy_0} \over g}\]Let's recall that \(R = \dot x_0 t\), let's substitute in t\[R = v_0 \cos(\theta) \left [ - a + \sqrt{v_0^2(\sin(2\theta) + 1) + 2gy_0} \over g \right]\]Things are a bit ugly here. Let's try to simplify it\[R = {-(v_0^2 \left ({1 \over 2} \sin(2 \theta) + \cos^2(\theta) \right)) + v_0 \cos(\theta) \sqrt{v_0^2 (\sin(2\theta) + 1) + 2 gy_0} \over g}\]Now, I would say take the derivative with respect to \(\theta\), set R =0 and solve for \(\theta\), but that is going to be one heck of a derivative.

This solution is not very graceful. Let me see if I can come up with a solution by using conservation of energy.

Answer 4

Perhaps we should rotate our coordinate plane 45 degrees and align it with the slope and vectorize gravity. That might eliminate several trigonometric terms.

Answer 5

I'll work on that one too.

Answer 6

@eashmore are you still there?

Answer 7

Yep. Sorry about the sloppy solution. I'm trying to think of the best way to reapproach this.

Answer 8

@expermentX....there is another interesting problem

Answer 9

using newton's method to solve the problem

Answer 10

Wow. Thanks for putting so much effort in @phi

It appears as if you arrived at an equally un-elegant solution in terms of the number of trigonometric terms. This is definitely a doozy.

Answer 11

using newton's method (fixed definition of constant k=1 + 64h/v^2)

If there is an "elegant" solution, I would like to see it. But I have a hunch this is as pretty as it gets.