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:
[tex]s = ut + \frac{1}{2} a t^2[/tex]
[tex]18 = 2 \times 3.3 + \frac{1}{2}a \times 3.3^2[/tex]
[tex]a = \frac{11.4 \times 2}{10.89}[/tex]
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 = [tex]\frac{2.09}{9.8}[/tex]
Θ = 12.31°
Hence, the angle of the inclined plane is: Θ = 12.31°
