A boy on skates accelerates to a speed of 11 m / s and rolls onto an ice slide

A boy on skates accelerates to a speed of 11 m / s and rolls onto an ice slide. But how high can he rise if the coefficient of friction is 0.1, and the angle of inclination of the hill to the horizon is 45 degrees.

V0 = 11 m / s.

μ = 0.1.

g = 9.8 m / s ^ 2.

∠α = 45 “.

h -?

Let’s write 2 Newton’s law for the projections of forces on an inclined plane: m * a = Ftr + m * g * sinα;

N = m * g * cosα.

The friction force is determined by the formula: Ftr = μ * N.

m * a = μ * m * g * cosα + m * g * sinα;

a = μ * g * cosα + g * sinα = g * (μ * cosα + sinα);

The movement of the boy S along the inclined plane is found by the formula: S = V0 ^ 2/2 * a = V0 ^ 2/2 * g * (μ * cosα + sinα).

The height h the boy will climb will be determined by the formula: h = S * sinα = V0 ^ 2 * sinα / 2 * g * (μ * cosα + sinα).

h = (11 m / s) ^ 2 * 0.7 / 2 * 9.8 m / s ^ 2 (0.1 * 0.7 + 0.7) = 5.6 m.

Answer: the boy will rise to a height of h = 5.6 m.