Physics Help

[minime72706]

Thanks too all who helped, or tried too.

 

[TuxDave]

Originally posted by: minime72706
Thanks too all who helped, or tried too.

No prob... I hope you got the answer you were looking for.

 

[DrPizza]

LMAO...
I just glanced at the paper plate that I wrote on and realized the simplest solution is sitting right there...
nearly the same solution as well, except without realizing the monkey and hunter problem (apple and bullet, whatever you want to call it)

d=vt
140 = V cos35 * t
V = 140/(t*cos35)

Now,
21 = V*sin35*t - 1/2 (9.81) * t^2
2 equations.... solve for t...
substitute for V
21 = 140*sin35 * t / (t * cos35) - 1/2(9.81)*t^2
a little cancelling (how'd I miss it the first time!!)
Ans: I realized there were 2 solutions and this would only give one of the solutions
21 = 140 tan 35 - 1/2 (9.81)*t^2
t = 3.9628s
(Same time as TuxDave)

140m/3.9628s = 35 m/s horizontal velocity... = V cos 35
divide by cos 35 and V = 43.1 m/s

Same answer

 

[DrPizza]

Now.... why didn't my solution give both possible answers? There should be a solution for on the way up as well as on the way down (except that it has to land on the wall, so we assume it's on the way down)

 

[TuxDave]

Originally posted by: DrPizza
LMAO...
I just glanced at the paper plate that I wrote on and realized the simplest solution is sitting right there...
nearly the same solution as well, except without realizing the monkey and hunter problem (apple and bullet, whatever you want to call it)

Yeah, I couldn't recall what it was called and so I just chose apple and bullet. I figure an apple/bullet would avoid any problems from PETA supporters that might not like the fact that I'm shooting monkeys out of trees. Nice job.

 

[TuxDave]

Originally posted by: DrPizza
Now.... why didn't my solution give both possible answers? There should be a solution for on the way up as well as on the way down (except that it has to land on the wall, so we assume it's on the way down)

Yeah, I was asking myself that. The solution we give doesn't constrain it to be on the way up or down so perhaps in some cases, our solution will be hitting the fence on the way up. However, I'm trying to find an intuitive reason such that if there is a solution on the way down, maybe there's no solution on the way up and vice-versa.

 

[DrPizza]

wow... I still can't come up with an intuitive way to show that there's only 1 solution. I know now that there is only one... but, it defied my initial logic.

 

[DrPizza]

Got it:
look at tuxdave's picture (fizicks)

For it to hit on the way up, it'd have to obviously have a higher initial velocity...
but, if it had a higher initial velocity, it would deviate from the hypotenuse less.

Or, think of a continuous function for velocity that can be used to visualize this:
instead of a ball, let's shoot a stream of water.
At a very low velocity, the water just barely trickles out... but as we increase the velocity of the water by turning the faucet on more, the water hits farther and farther away (hold the 35degree angle constant).... There's only 1 velocity where the water will hit the top of the fence.. after that, when the velocity is higher, it will be going over the fence.

 

[TuxDave]

Originally posted by: DrPizza
Got it:
look at tuxdave's picture (fizicks)

For it to hit on the way up, it'd have to obviously have a higher initial velocity...
but, if it had a higher initial velocity, it would deviate from the hypotenuse less.

Or, think of a continuous function for velocity that can be used to visualize this:
instead of a ball, let's shoot a stream of water.
At a very low velocity, the water just barely trickles out... but as we increase the velocity of the water by turning the faucet on more, the water hits farther and farther away (hold the 35degree angle constant).... There's only 1 velocity where the water will hit the top of the fence.. after that, when the velocity is higher, it will be going over the fence.

Heh... I had a less intuitive explaination. Thanks.

My way of seeing it was since the path of the ball follows a parabolic path (2nd order quadratic), we're given 2 points (the starting point and the end point), and the derivative at the starting point (the 35 degree angle). Somehow I convinced myself that those three pieces of information is enough to define a parabola and hence only have one solution.