Math  /  Trigonometry

Question39. (II) A skateboarder, with an initial speed of 2.0 m/s2.0 \mathrm{~m} / \mathrm{s}, rolls virtually friction free down a straight incline of length 18 m in 3.3 s . At what angle θ\theta is the incline oriented above the horizontal?

Studdy Solution
Use the acceleration to find the angle of the incline. The acceleration down the incline is due to gravity and can be expressed as:
a=gsin(θ) a = g \cdot \sin(\theta)
where g=9.8m/s2 g = 9.8 \, \text{m/s}^2 is the acceleration due to gravity. Solve for θ \theta :
sin(θ)=ag=2.099.8 \sin(\theta) = \frac{a}{g} = \frac{2.09}{9.8}
θ=arcsin(2.099.8) \theta = \arcsin\left(\frac{2.09}{9.8}\right)
Calculate θ \theta :
θarcsin(0.213)12.3 \theta \approx \arcsin(0.213) \approx 12.3^\circ
The angle of the incline is approximately:
12.3 \boxed{12.3^\circ}

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