contestada

a 74 kg man jumps from a window 1.6 m above a sidewalk. the acceleration of gravity is 9.81 m/s 2 . a) what is his speed just before his feet strike the pavement? answer in units of m/s

Respuesta :

sqdancefan

Answer:

  5.60 m/s

Explanation:

His initial potential energy is equal to his final kinetic energy.

  mgh = 1/2mv^2

  2gh = v^2 . . . . . . . multiply by 2/m

  v = √(2gh) = √(2(9.81 m/s²)(1.6 m)) = √(31.392 m²/s²)

  v ≈ 5.60 m/s

_____

Additional comment

The mass is irrelevant in the equations we typically use for this purpose.

Otras preguntas