Answer:
g= 1.8 m/s²
Explanation:
From conservation of energy we know:
[tex]mgh=\frac{1}{2}mv^2[/tex]
Solving for g with h being the height of the ramp with angle ∅ and length L:
(1) [tex]g=\frac{v^2}{2h}=\frac{v^2}{2Lsin\theta}[/tex]
The velocity v at the bottom of the ramp for a constant acceleration a in a time t is given by:
(2)[tex]v=at=gsin\theta t[/tex]
Plugging equation 2 in to equation 1:
(3) [tex]g=\frac{2L}{sin\theta t^2}[/tex]