The stone is thrown vertically upward with an initial speed of 25 m / s.

The stone is thrown vertically upward with an initial speed of 25 m / s. Determine the speed module of the stone in 1.5 s after the start of the movement

V0 = 25 m / s.

g = 10 m / s2.

t = 1.5 s.

V -?

If we neglect the force of air resistance, only the gravity force m * g, directed in the opposite direction of its movement, acts on the stone during movement. Therefore, it will decelerate with the acceleration of gravity g.

When braking with constant acceleration g, the speed of the body changes according to the formula: V = V0 – g * t, where V0 is the speed of the stone when throwing.

V = 25 m / s – 10 m / s2 * 1.5 s = 10 m / s.

Answer: the speed of the stone after a time t = 1.5 from the start of the throw will be V = 10 m / s.