From the steep bank of the river 20 m high, a stone was thrown in a horizontal direction at a speed of 15 m / s.

From the steep bank of the river 20 m high, a stone was thrown in a horizontal direction at a speed of 15 m / s. After what time, at what speed and at what angle will the stone fall into the water?

h = 20 m.

g = 9.8 m / s ^ 2.

Vх = 15 m / s.

t -?

V -?

∠α -?

h = V0y * t + g * t ^ 2/2. Since V0у – 0, the formula will take the form: h = g * t ^ 2/2.

t = √ (2 * h / g).

t = √ (2 * 20 m / 9.8 m / s ^ 2) = 2.02 s.

The horizontal component at the moment of impact will be Vx = 15 m / s, since the stone moves horizontally evenly.

The vertical component of the velocity Vу at the moment of impact on the water is found by the formula: Vу = g * t.

Vу = 9.8 m / s ^ 2 * 2.02 s = 19.8 m / s.

V = √ (Vy ^ 2 + Vx ^ 2).

V = √ ((19.8 m / s) ^ 2 + (15 m / s) ^ 2) = 24.8 m / s.

tgα = Vх / Vу.

∠α = arctan Vх / Vу.

∠α = arctan (15 m / s / 19.8 m / s) = 36.8 “.

Answer: t = 2.02 s, V = 24.8 m / s, ∠α = 36.8 “.