# The skier drives off the top of the hill with an acceleration of 2.3 m / s².

**The skier drives off the top of the hill with an acceleration of 2.3 m / s². The height of the hill is 180 m, and its slope is 30 ° (degrees). Find the descent time.**

a = 2.3 m / s2.

h = 180 m.

V0 = 0 m / s.

∠α = 30 °.

t -?

With uniformly accelerated movement, the path S traversed by the body is determined by the formula: S = V0 * t + a * t ^ 2/2, where V0 is the initial speed of movement, t is the time of movement, a is the acceleration during movement.

Since the initial speed of the skier during the descent is V0 = 0 m / s, the formula will take the form: S = a * t ^ 2/2.

Let us express the length of the descent S as the hypotenuse of a right-angled triangle with the opposite leg h = 180 m and ∠α = 30 °.

sinα = h / S.

S = h / sinα.

h / sinα = a * t ^ 2/2.

t = √ (2 * h / a * sinα).

t = √ (2 * 180 m / 2.3 m / s2 * sin30 °) = 17.7 s.

Answer: the time of the skier’s descent from the mountain was t = 17.7 s.