Determine the acceleration of a sled rolling down a hill with a height of h = 2m

Determine the acceleration of a sled rolling down a hill with a height of h = 2m and a length of l = 4m. Neglect friction and air resistance.

The acceleration acquired by the body as a result of the impact on it is directly proportional to the strength of this impact and inversely proportional to the body mass:
If several forces act on the body, then their resultant is taken.
F = m * a
In our case, the following forces act on the body:
Ftre + N + P = m * a, where Ffre is the friction force (by the condition we will neglect it), N is the reaction of support to the body weight, and is equal to it in modulus and opposite to the direction, P is the body weight.
Let us draw the axes with the condition that the X-axis is directed in the direction of motion of the body, and the Y-axis is in the direction of the support reaction.
The body moves at an angle α to the horizon, and sin α = h / l
Let’s write in projections on the axis:
ОХ: P * sin α = ma
OU: N-P * cos α = 0
We are interested in the expression:
P * sin α = ma, where – P – body weight P = m * g
With this in mind, we get:
m * g * sin α = ma
Let us express the acceleration from this expression:
a = g * sin α = 9.8 * h / l = 9.8 * 2/4 = 4.9 m / s²
Answer: body acceleration 4.9 m / s².