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How fast would a penny be falling when it reaches ground level if dropped from rest off the 102nd floor observation deck of the empire state building? (381 meters above the ground) You may treat air resistance as negligible.

Respuesta :

Air resistance is negligible

  • Distance=381m=s
  • Acceleration due to gravity=g=10m/s^2
  • Initial velocity=u=0m/s
  • Final velocity=v=?

Apply third equation of kinematics

[tex]\\ \tt\longmapsto v^2=u^2+2gs[/tex]

  • u=0

[tex]\\ \tt\longmapsto v^2=2(10)381[/tex]

[tex]\\ \tt\longmapsto v^2=7620[/tex]

[tex]\\ \tt\longmapsto v=87.29ms^{-1}[/tex]

Answer:

According to the Given Statement

Penny is dropped from rest of the 102nd floor which is 381 metres above from the ground.

we 'll take here the acceleration due to gravity (g ) is 9.8m/s²

Its initial velocity u be 0 m/s ( as it is at rest)

Using Kinematic Equation

  • v² = u² + 2gh

here h is the height of 102nd floor

➥ v² = 0² + 2 × 9.8 × 381

➥ v² = 0 + 19.6 × 381

➥ v² = 7467.6

➥ v = √7467.6

➥ v = 86.41 m/s

So, 86.41 m/s penny would be fall.