A stone is thrown vertically upward with a speed of 14.6 m/s from the edge of a cliff 93 m high. What total distance did it travel

Respuesta :

Answer:

The  value is [tex]H = 114.74 \ m[/tex]

Explanation:

From the question we are told that  

  The initial  speed is  [tex]u = 14.6 \ m/s[/tex]

   The height of the cliff is  [tex]h_o = 93 \ m[/tex]

Generally from kinematic equation we have that

       [tex]v^2 = u^2 + 2as[/tex]

At maximum height , v which is the finial velocity is zero

So

     [tex]0^2 = 14.6^2 + 2 * (- 9.8) * s[/tex]

=>  [tex]s = 10.87 \ m[/tex]

Generally the total distance traveled is mathematically represented as

      [tex]H = 2 * s + h_o[/tex]

=>   [tex]H = 2 * 10.87 + 93[/tex]

=>   [tex]H = 114.74 \ m[/tex]

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