Question

Please I need help with a new question, this has a solution that I can't still dig. See the attached JPG

Answer 1

This is the solution I cant dig. Any help is welcome.

Answer 2

The equations for motion under constant acceleration are:
\[v=u+at\]
\[s=ut+at ^{2}/2\]
In this case u=0. And the question acts for motion ALONG the board. But the given acceleration is vertical. You've to break it into its components, such that one component is along the board.

So the acceleration along the board is a0sintheta. Now just substitute:
\[a=a0\sin \theta\]