dayanaratorres1066 dayanaratorres1066
  • 08-02-2020
  • Physics
contestada

A 120 kg base jumper stands on a bridge at an elevation of 250 meters above the water. To the nearest tenth, what is her speed just before she strikes the water

Respuesta :

joseaaronlara
joseaaronlara joseaaronlara
  • 08-02-2020

Answer:

The answer to your question is vf = 70 m/s

Explanation:

This is a free fall problem and in these kinds of problems, the mass is not important to calculate the speed.

Data

mass = 120 kg

height = 250 m

g = 9.81 m/s²

Formula

  h = [tex]\frac{vf^{2}}{2g}[/tex]

Solve for vf

   vf² = 2gh

   vf = [tex]\sqrt{2gh}[/tex]

Substitution

   vf = [tex]\sqrt{2(9.81)(250)}[/tex]

Simplification

    vf = [tex]\sqrt{4905}[/tex]

Result

   vf = 70 m/s

Answer Link

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