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

The angle at which the face of the wave is inclined above the horizontal is approximately 21.5°.

Step-by-step explanation:

The surfer's acceleration is 3.50 m/s² and friction can be ignored. To find the angle at which the face of the wave is inclined above the horizontal, we can use the formula:

a|| = g * sin(θ)

Where a|| is the acceleration parallel to the incline, g is the acceleration due to gravity (approximately 9.8 m/s²), and θ is the angle of the incline.

In this case, a|| = 3.50 m/s². Rearranging the formula, we get:

sin(θ) = a|| / g

Substituting the given values, we have:

sin(θ) = 3.50 / 9.8

θ = arcsin(3.50 / 9.8)

Using a calculator, we find that θ is approximately 21.5°.

Answer 2

`\theta=20.92^(\circ)`

Step-by-step 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)((a)/(g))`

`\theta=sin^(-1)((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.