Answer:
[tex]\theta=20.92^{\circ}[/tex]
Explanation:
Given that,
Acceleration of the surfer, [tex]a=3.5\ m/s^2[/tex]
To find,
The angle.
Solution,
Let [tex]\theta[/tex] is the angle at which the face of the wave is inclined above the horizontal. By considering the free body diagram of the inclined plane,
[tex]ma=mg\ sin\theta[/tex]
[tex]\theta=sin^{-1}(\dfrac{a}{g})[/tex]
[tex]\theta=sin^{-1}(\dfrac{3.5}{9.8})[/tex]
[tex]\theta=20.92^{\circ}[/tex]
Therefore, the angle at which the face of the wave is inclined above the horizontal is 20.92 degrees.
