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Se lanza una pelota de béisbol desde la azotea de un edificio de 25 m de altura con velocidad inicial de magnitud 10 m/s y dirigida con un ángulo de 63.1° sobre la horizontal. A) ¿Qué rapidez tiene la pelota justo antes de tocar el suelo? Use métodos de energía y desprecie la resistencia del aire.

Sagot :

Answer:

 v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

Starting point. On the roof of the building

         Em₀ = K + U = ½ m v₀² + m g y₀

Final point. On the floor

         Em_f = K = ½ m v_f²

         Emo = Em_g

         ½ m v₀² + m g y₀ = ½ m v_f²

        v_f² = v₀² + 2 g y₀

         

let's calculate

        v_f = √(10² + 2 9.8 25)

        v_f = 24.3 m / s

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