Question

A skateboarder, with an initial speed of 2.0 m/s, rolls virtually friction free down a straight incline of length 18 m in 3.3 s. at what angle θ is the incline oriented above the horizontal?

Answer 1

Initial speed of the skateboarder (u) = 2 m/s

Distance covered (s) = 18 m

Time taken = 3.3 seconds

Let the acceleration be a.

Using seconds equation of motion:

$s = ut + \frac{1}{2} a t^2$

$18 = 2 \times 3.3 + \frac{1}{2}a \times 3.3^2$

$a = \frac{11.4 \times 2}{10.89}$

a = 2.09 m/s^2

Now, Acceleration down the incline = g Sin Θ

g Sin Θ = a

9.8 × Sin Θ = 2.09

Sin Θ = $\frac{2.09}{9.8}$

Θ = 12.31°

Hence, the angle of the inclined plane is: Θ = 12.31°