Answer:
a) y₁ = ½ a, b) y₂ = 4 y₁, c) y₃ = 9 y₁
Explanation:
For this exercise we can use the accelerated motion relationships.
Let's set a reference system where the x axis is parallel to the plane and its positive side is going down the plane.
y = y₀ + v₀ t + ½ a t²
in that case where we throw the marble is the zero point, y₀ = 0, as part of rest its initial velocity is zero v₀ = 0 and a is the acceleration along the inclined plane
y = ½ a t²
a) in the first second t = 1
y₁ = ½ a
b) in the next second of movement
t = 2 s
y₂ = ½ a 2²
y₂ = 4 ½ a
y₂ = 4 y₁
c) for the next second
t = 3 s
y₃ = ½ a 3²
y₃ = 9 ½ a
y₃ = 9 y₁