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A 754 N diver drops from a board 9.80 m above the water’s surface. The acceleration of gravity is 9.81 m/s 2 . a) Find the diver’s speed 4.70 m above the water’s surface. Answer in units of m/s.

Respuesta :

Answer:

[tex]v=10.0031\ m.s^{-1}[/tex]

Explanation:

Given:

  • weight of the diver, [tex]w=754\ N[/tex]
  • height of descend, [tex]h'=9.8\ m[/tex]
  • height of observation, [tex]h=4.7\ m[/tex]

Displacement of the diver:

[tex]s=h'-h[/tex]

[tex]s=9.8-4.7[/tex]

[tex]s=5.1\ m[/tex]

Now using the equation of motion:

[tex]v^2=u^2+2g.s[/tex]

where;

[tex]v=[/tex] final velocity of the diver at the observed time

[tex]u=[/tex] initial velocity of the diver when at the top height = 0

[tex]v^2=0^2+2\times 9.81\times 5.1[/tex]

[tex]v=10.0031\ m.s^{-1}[/tex]

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