How Does Vertical Distance y(t) Satisfy the Given Differential Equation?

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SUMMARY

The discussion centers on the differential equation governing the vertical distance y(t) traveled by an object on an inclined plane, expressed as dy/dt = sqrt(2g/(1+I*)) * (sin alpha) * sqrt(y). The equation relates the vertical distance to gravitational acceleration g, the angle of inclination alpha, and a term I* that requires clarification. Participants emphasize the importance of understanding the relationship between velocity and the derivative of position, while also noting the need for context from Problem 1 to fully grasp the concepts involved.

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2. If y(t) is the vertical distance traveled at time t, then the same reasoning as used in Problem 1shows that v2= (2gy)/(1+I*) at any time. Use this result to show that y satisfies the differential equation

dy/dt=sqrt(2g/(1+I*))*(sin alpha)*sqrt(y)

where a is the angle of inclination of the plane.

I plugged dy/dt into v, because is the derivative of position. I have dy/dt=sqrt(2g/(1+I*))*sqrt(y) because you can pull it out, but i don't know how to get the sin alpha
 
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It would have been better to include problem "1" rather than expecting people to gues what the problem really is.

It looks like you have an object moving on an inclined plane and y is the vertical height. But what is I*?
 

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