The skater, having accelerated, drives onto an ice slide, inclined at an angle of 30 ° to the horizon, and passes 10 m

The skater, having accelerated, drives onto an ice slide, inclined at an angle of 30 ° to the horizon, and passes 10 m to a full stop. What was the speed of the skaters before starting the ascent?

Given:

a = 30 degrees – the angle of inclination of the slide to the horizon;

g = 10 m / s ^ 2 – acceleration of gravity.

S = 10 meters – the path that the skater goes to a complete stop.

It is required to find v0 (m / s) – the initial speed of the skater before going up the hill.

Since, according to the condition of the problem, it is not indicated, the friction force acting on the skater is not taken into account. Then, according to Newton’s second law, the acceleration is:

m * g * sin (a) = m * a

a = g * sin (a).

V0 = (2 * S * a) ^ 0.5 = (2 * S * g * sin (a)) ^ 0.5 = (2 * 10 * 10 * 0.5) ^ 0.5 = (100) ^ 0.5 = 10 m / s.

Answer: the speed of the skater before climbing the hill was 10 m / s.