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I'm trying to help someone with a physics problem. So this is what he described:

"(01:01:49) ka0s1337the0ry: if she gives the puck a velocity of 4.6m/s along the length (1.75m) of the table at one end

(01:01:53) RAMEN 3bun BR: are you going to be a freshman in high school?

(01:02:06) ka0s1337the0ry: by the time it reached the other end the puck has drifted 3.60 cm to the right

(01:02:13) B0x0Rr0X0r402: vectors

(01:02:14) B0x0Rr0X0r402: etc

(01:02:17) ka0s1337the0ry: but it still has a velocity component along the length of 4.60m/s.

(01:02:32) ka0s1337the0ry: She conludes correctly that the table is not level and correctly calculates its inclination from the above information.

==

Okay - so we have the formula [tex]v_f^2 = v_i^2 + 2 a (x_f - x_i)[/tex] and F=ma.

He says that v_f = 0. Now we have [tex]\delta x = \sqrt{0.036^2+1.75^2}[/tex]

We now have [tex]a = \frac{-4.6^2}{2 \sqrt{0.036^2+1.75^2}} = -6.16 m / s^2[/tex]

Then we use [tex]F = ma[/tex]. Since gravity is acting against the object, [tex]mg \sin \theta = ma[/tex]. Since we want theta, we can have [tex]\theta = \arcsin{(\frac{ma}{mg})[/tex] We get [tex]\theta = -38.08[/tex], which is effectively 38.08 degrees. Is my reasoning correct? He said that most of his classmates got values around 2 degrees

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# Homework Help: Puck on table problem

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