Question

A skateboarder, with an initial speed of 4.0 m/s , rolls virtually friction free down a straight incline of length 19 m in 3.3 s .

At what angle θ is the incline oriented above the horizontal?

Express your answer using two significant figures.

Answer 1

19 m in 3.3 s

That means, will be, = d/t

=19/3.3

=5.76 m/s at the bottom of the ramp.
Use v² = v₀² + 2ax

to find the lateral acceleration. Solving for a, we get

a = (v² - v₀²)/(2x).
v = 5.76 m/s; v₀ = 4 m/s; x = 19 m. Substituting, we get

a = (5.76²-4²)/(2*19)

a=0.452m/s²

This acceleration equals g·sin(θ);

,therefore θ = sinˉ¹(a/g)

= sinˉ¹((0.452 m/s²)/(9.81 m/s²))

= 2.64°.