A body weighing 1 kg, thrown vertically upward at a speed of 40 m / s, reached its highest point in 2.5 s.

A body weighing 1 kg, thrown vertically upward at a speed of 40 m / s, reached its highest point in 2.5 s. Find the value of the force of air resistance, considering it constant.

m = 1 kg.

V0 = 40 m / s.

V = 0 m / s.

g = 10 m / s ^ 2.

t = 2.5 s.

Fsopr -?

The acceleration of the body, and we find it by the formula: a = (V – V0) / t. Since V = 0 m / s, then a = – V0 / t.

The sign “-” means that the acceleration is directed in the opposite direction to the movement, the body is inhibited.

Let us write Newton’s 2 law for a body when lifting in vector form: m * a = m * g + Fcopr, where m is the body mass, a is the acceleration of the body’s motion, m * g is the force of gravity, Fcopr is the resistance to motion.

Fcopr = m * a – m * g.

Fcopr = m * (a – g).

Fcopr = m * (V0 / t – g).

Fcopr = 1 kg * (40 m / s / 2.5 s – 10 m / s ^ 2) = 6 N.

Answer: the resistance force is Fcopr = 6 N.