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

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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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