Question

A surfer "hangs ten", and accelerates down the sloping face of a wave. If the surfer’s acceleration is 3.50 m/s2 and friction can be ignored, what is the angle at which the face of the wave is inclined above the horizontal?

Answer 1

Answer:

$\theta=20.92^{\circ}$

Explanation:

Given that,

Acceleration of the surfer, $a=3.5\ m/s^2$

To find,

The angle.

Solution,

Let $\theta$ 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,

$ma=mg\ sin\theta$

$\theta=sin^{-1}(\dfrac{a}{g})$

$\theta=sin^{-1}(\dfrac{3.5}{9.8})$

$\theta=20.92^{\circ}$

Therefore, the angle at which the face of the wave is inclined above the horizontal is 20.92 degrees.