Just wanting to double check the reasoning and the work for these. They are fairly simple, so I didn't use latex, but I can re-edit at any time if someone so desires (I'm online more often than I like to admit hehehe).(adsbygoogle = window.adsbygoogle || []).push({});

Thanks.

1. The problem statement, all variables and given/known data

1) A mass 'm' is rolled of a table at height 'h' with horizontal speed 'Vx'. Where does the mass land? What trajectory did the mass take?

2) If x = -cos(t) + 3*sin(t), what is the amplitude and phase?

3) Show that an equivalent expression for x = C1 * cos(wt) + C2 * sin(wt) is B*cos(wt + P). How do B and P depend on C1 and C2.

2. Relevant equations

1) F = ma

2) sin(a +- b) = sin(a)cos(b) +- cos(a)sin(b)

3) cos(a +- b) = cos(a)cos(b) -+ sin(a)sin(b)

3. The attempt at a solution

1) I solved the equations of motion for both vertical and horizontal directions.

With y(0) = h and Vy(0) = 0, I got y(t) = h - mg/2 * t^2

With x(0) = 0 and Vx(0) = Vx, I got x(t) = Vx * t

Solving for y = 0, t_land = sqrt(2/[mg]), so I said the mass will have fallen and displaced Vx * t_land away from table.

Plugging x/Vx for 't', I have that the path is y(t) = h - mg/2 * (x/Vx)^2, which is a parabola with downward concavity.

2) If we rewrite as A * sin(wt + P), then A * sin(P) = C1 and A * cos(P) = C2

Therefore A is simply sqrt(C1^2 + C^2) = sqrt(10).

and P = arctan(C1/C2) = arctan(-1/3).

3) If we rewrite as B * cos(wt + P), then B * cos(P) = C1 and - B * sin(P) = C2

Therefore B is simply sqrt(C1^2 + C^2)

and P = arctan(-C2/C1)

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# Homework Help: Checking very simple problems of motion

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