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

Answer 2

Lol, isn't this physics stuff? Well, no matter, you want to go through the answer step by step?

Answer 3

yes, please! And I posted it in physics. But it is also math.

Answer 4

Okay, Do you know your formula for projectile motion?

Answer 5

I know:\[x_t = x_0 + v_x\]and that \[v_x = 20 \cos \theta\]And \[y_t = y_0 + v_y -16t^2\]and that \[v_y = 20 \sin \theta\]

Answer 6

Ok, the first question asks how to maximize the distance... does your instruction specify horizontal distance or just distance?

Answer 7

Oh ok, I saw from the diagram...

Answer 8

So if we aim to maximize, we would do well to have an equation to work with, can you come up with one?

Answer 9

no, that is where I'm stuck.

Answer 10

I see, well, how do you calculate, for instance, the distance, if you already know how far it traveled horizontally and vertically?

Answer 11

I really don't know. I can use the formulas, but...

Answer 12

All right, give it a good think, and I'll do the same

Answer 13

Not too sure about this, but could it be that it requires a rotation of axes?

Answer 14

hey, @BTaylor, are you still here? :)

Answer 15

If we create a graph of height as a function of time (y = -16t^2 + 20tcos(theta) + 6), it would "hit the ground" when it intersects the line y = -x. So we could add x, to get
y = -16t^2 + 20tcos(theta) + 6 + t
when y = 0, the pit is on the ground.

Answer 16

I never would have thought of that... I guess my creativity is sorely lacking.

Answer 17

t= time
A= angle

The vertical distance of the pit as a function of time:
x(t) = 20cos(A)t

And the height of the ground at any given x is:
h(x) = -x
since the hill is a 45 degree angle.

Since we have x as a function of time, height as a function of time is:
h(x(t)) = -20cos(A)t

The height of the pit is:
y(t) = 6 + 20sin(A)t - 16t^2

And the pit will land when that height equals the height of the hill.
y(t) = h(t)
6 + 20sin(A)t - 16t^2 = -20cos(A)t

Answer 18

First thing should have said "Horizontal distance of the pit..."

Answer 19

This is making me realize I need to brush up on my physics, I'm feeling completely blank here o.o

Answer 20

6+20[sin(A)+cos(A)]t -16t^2 = 0

Answer 21

This problem is so messy... I feel like there's a simpler way to solve it.

Answer 22

could I graph 6+20[sin(A)+cos(A)]t -16t^2 = 0 with x = A and y = t ?

Answer 23

The best thing I can think to do from here is use quadratic formula to get
t in terms of A
Then you can use
x(t) and y(t), substitute to get
x(A) and y(A)

Answer 24

I posted this in physics. Does that answer help?
http://openstudy.com/study#/updates/4fdf5b55e4b0f2662fd5aed3

Answer 25

\[\large \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
\[\large \frac{-20(cosA+sinA) \pm \sqrt{(20(sinA+cosA))^2 - 4*6*-16}}{2*6}\]

Answer 26

@SmoothMath wouldn't it be 2 x -16 as the denominator?

Answer 27

Yes. Thank you.

Answer 28

\[= \frac{-20(sinA+cosA) \pm \sqrt{400(1+2sinAcosA) + 384}}{32}\]

Answer 29

\[= \frac{-20(sinA + cosA)\pm \sqrt{784 + 800sinAcosA}}{-32}\]

Answer 30

so...now what?

Answer 31

Well. That's t in terms of A. And we have x(t) = 20cos(A)*t
so
\[x(A) = 20\cos(A)*\frac{-20(sinA + cosA)\pm \sqrt{784 + 800sinAcosA}}{-32}\]
If we derive and maximize that, then we can just use pythagorean theorem on that 45-45-90 triangle and we'll have it.

Of course, maximizing that is a pain, so I'm gonna try letting Alpha do it for me.

Answer 32

this seems most easily solved using a numerically approach. For comparison, I got A= 20.18266752 degrees,t= 1.811522804 seconds, and measured distance d = 48.09148

Answer 33

how?