vegeta7722
vegeta7722 vegeta7722
  • 24-10-2017
  • Physics
contestada

An object is dropped from a 15 m ledge. How fast it is moving just before it hits the ground?

Respuesta :

uguerra2003 uguerra2003
  • 24-10-2017
By v^2 = u^2 + 2gh 
v^2 = 0 + 2 x 9.8 x 15
v = √294
v = 17.15 m/s 
Answer Link
aristocles aristocles
  • 15-10-2019

Answer:

[tex]v_f = 17.15 m/s[/tex]

Explanation:

Since the ball is dropped from the height

h = 15 m

so here initial speed of the ball is

[tex]u = 0[/tex]

acceleration of the ball due to gravity is given as

[tex]a = 9.8 m/s^2[/tex]

now the final speed of the ball is given as

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]v_f^2 - 0 = 2(9.8)(15)[/tex]

[tex]v_f = 17.15 m/s[/tex]

Answer Link

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