The car with the engine turned off drove 50 m down the road at an angle of 30 degrees to the horizon.

The car with the engine turned off drove 50 m down the road at an angle of 30 degrees to the horizon. At the same time, its speed reached 30 ms. What is the initial vehicle speed? Friction neglected

∠α = 30 °.

S = 50 m.

g = 9.8 m / s2.

V = 30 m / s.

V0 -?

When the car moves on a plane, 2 forces act on it: the force of gravity m * g directed vertically downward, N is the reaction force of the plane directed perpendicular to the inclined plane.

m * a = m * g + N – 2 Newton’s law in vector form.

OH: m * a = m * g * sinα.

OU: 0 = – m * g * cosα + N.

N = m * g * cosα.

a = m * g * sinα / m = g * sinα.

We express the acceleration of the car by another formula: a = (V ^ 2 – V0 ^ 2) / 2 * S.

g * sinα = (V ^ 2 – V0 ^ 2) / 2 * S.

V ^ 2 – V0 ^ 2 = 2 * S * g * sinα.

V0 ^ 2 = V ^ 2 – 2 * S * g * sinα.

V0 = √ (V ^ 2 – 2 * S * g * sinα).

V0 = √ ((30 m / s) 2 – 2 * 50 m * 9.8 m / s2 * sin30 °) = 20 m / s.

Answer: the initial vehicle speed was V0 = 20 m / s