PWA "The Lounge"

[FenwickHockey65]

Originally Posted by DGthe3

Unknowable, unless you assume that the spider lands on the magazine at the apex of its trajectory.

But beyond that ... the answer I got was 19.22mm. Is that correct?

Nope.

[DGthe3]

Originally Posted by FenwickHockey65

Nope.

Alright, time to work backwards. What is the answer?

[Da Hose]

Originally Posted by DGthe3

7.94mm is roughly 5/16 of an inch

Ok feasible.... I must be getting tired!!!! I read mm. It registered cm
Time for da hose to hit da hay!!!!

[sugar'n'spice]

Originally Posted by Da Hose

Ok feasible.... I must be getting tired!!!! I read mm. It registered cm
Time for da hose to hit da hay!!!!

Good night, Ronnie!!!


Me too!!!

[Da Hose]

Originally Posted by sugar'n'spice

Good night, Ronnie!!!


Me too!!!

Nite Jo!!!!!!

[FenwickHockey65]

Originally Posted by DGthe3

Alright, time to work backwards. What is the answer?

No idea. It's an online homework thing, I put answers in and it tells me if they're right or not.

[GTAHVIT]

Night gang!

[Da Hose]

Originally Posted by GTAHVIT

Night gang!

Nite bro!

[sugar'n'spice]

Originally Posted by GTAHVIT

Night gang!

Nighty night!!!

["prima"Donna]

Originally Posted by Da Hose

Woo hoo..... Insert donna's happy dance here!!!!!!
Lmao...... iPhone posting has it's disadvantages

LOL, happy to oblige!!!

[DGthe3]

Originally Posted by FenwickHockey65

No idea. It's an online homework thing, I put answers in and it tells me if they're right or not.

that sucks

[Xanthos]

Originally Posted by FenwickHockey65

WTF.

A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.850 m/s at an angle of 33.9° above the table, and it lands on the magazine 0.0740 s after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.

.85sin(33.9) = 0.47 m/s = Vertical velocity

Y=Vy*T + 1/2at^2

Y = .00794

= 7.94 mm

Someone tell me if I'm screwing up because I have no idea why the hell this is incorrect.

You're screwing up.

Your math is good but you're looking at it the wrong way. You have time and a constant x velocity, so the first thing you should solve for is that X velocity. Next, assume there is no magazine and solve for the total time it would take for the spider to travel in a normal, parabolic arc. Then you can figure out the difference in X distance. Next, you solve for your hypotenuse using the .850 m/s and the time not traveled. Finally, solve for Y distance based on the other two sides of the triangle.

Vx = (.850)(cos(33.9)) = .705 m/s
Vy = (.850)(sin(33.9)) = .474 m/s

Ttot = 2(Vy) = 2(.474) = .097 s
..........AG...........9.8

Ttot - Tact = Tlost = .097 - .074 = .023 s

dx = (.023)(.706) = .016 m
dh = (.023)(.850) = .020 m
dy = (dh^2-dx^2)^1/2 = .0256 m or 25.6 mm

Let me know if this helps.
- X

[DGthe3]

Originally Posted by Xanthos

You're screwing up.

Your math is good but you're looking at it the wrong way. You have time and a constant x velocity, so the first thing you should solve for is that X velocity. Next, assume there is no magazine and solve for the total time it would take for the spider to travel in a normal, parabolic arc. Then you can figure out the difference in X distance. Next, you solve for your hypontenuse using the .850 m/s and the time not traveled. Finally, solve for Y distance based on the other two sides of the triangle.

Vx = (.850)(cos(33.9)) = .705 m/s
Vy = (.850)(sin(33.9)) = .474 m/s

Ttot = 2(Vy) = 2(.474) = .097 s
..........AG...........9.8

Ttot - Tact = Tlost = .097 - .074 = .023 s

dx = (.023)(.706) = .016 m
dh = (.023)(.850) = .020 m
dy = (dh^2-dx^2)^1/2 = .0256 m or 25.6 mm

Let me know if this helps.
- X

that would work. Its been too long since I've solved any projectile motion stuff.

[Xanthos]

Originally Posted by DGthe3

that would work. Its been too long since I've solved any projectile motion stuff.

Lmao - its funny how easy it is to "overthink" the basics after doing more complicated stuff.
- X