A body thrown from a tower in a horizontal direction at a speed of 20

A body thrown from a tower in a horizontal direction at a speed of 20 m / s. Determine the speed and direction at the end of 2 seconds.

To determine the speed of a thrown body after 2 seconds of falling, we use the Pythagorean theorem: V = √ (Vv2 + Vg2) = √ ((g * t) 2 + Vg2).

Constants and variables: g – gravitational acceleration (g ≈ 10 m / s2); t is the elapsed time of the fall (t = 2 s); Vg – horizontal speed (Vg = 20 m / s).

Calculation: V = √ ((g * t) ^ 2 + Vg ^ 2) = √ ((10 * 2) ^ 2 + 20 ^ 2) = 28.28 m / s.

Since after 2 seconds Vg = Vw (vertical speed) = 20 m / s, the velocity vector at the considered moment of time is directed at an angle of 45º.

Answer: The speed of the thrown body is 28.28 m / s and is directed at an angle of 45º.