Abolt comes loose from underneath an elevator that is moving upward at a speed of 5 m/s. the bolt reaches the bottom of the elevator shaft in 3.1 s. (a) how high up was the elevator when the bolt came loose? (in m)
(b) what is the speed of the bolt when it hits the bottom of the shaft? (in m/s)

a)  The elevator was  31.64 m high up when the bolt came loose.

b)  Speed of the bolt when it hits the bottom of the shaft = 25.41 m/s

Explanation:

a) Considering motion of bolt:-

Initial velocity, u =  5 m/s

Acceleration , a = -9.81 m/s²

Time = 3.1 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= 5 x 3.1 - 0.5 x 9.81 x 3.1²

    s = 0 x t + 0.5 x 9.81 x t²

    s = -31.64 m

The elevator was  31.64 m high up when the bolt came loose.

b) We have equation of motion v = u + at

  Initial velocity, u =  5 m/s

 Acceleration , a = -9.81 m/s²

 Time = 3.1 s  

Substituting

  v = u + at

  v  = 5 - 9.81 x 3.1 = -25.41 m/s

Speed of the bolt when it hits the bottom of the shaft = 25.41 m/s

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