Integrate to find

Problem 40 finds the time T by integration.

For a vertical drop of $y$, equation (12) gives the velocity along the plane: $v^2=2gy/(1+J)$. Thus $v=c{y}^{1/2}$ for $c=\_\_\_\_\_$. The vertical velocity is $dy/dt=v\sin \alpha$:

$$dy/dt=c{y}^{1/2}\sin \alpha \text{and}\int y^{-1/2}dy=\int c\sin \alpha dt.$$

Integrate to find $y(t)$. Show that the bottom is reached ($y=h$) at time $T=2\sqrt{h}/c\sin \alpha$.

This exercise builds on another exercise in its source book, which isn't shown here.

Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §8.5 Masses and Moments, #40