Adiver runs horizontally off the end of a 3.0-m-high diving board with an initial speed of 1.8 m/s. given that the diver's initial position is xi=0 and yi=3.0m, find her x positions at the times t = 0.25 s, t = 0.50 s, and t = 0.75 s.

At t = 0.25 s,  x position = 0.45 m

At t = 0.50 s,  x position = 0.90 m

At t = 0.75 s,  x position = 1.35 m

Explanation:

At t = 0.25 s

Considering horizontal motion of diver:-

Initial velocity, u =  1.8 m/s

Acceleration , a = -0 m/s²

Displacement, s = ?

Time, t = 0.25 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

   s = 1.8 x 0.25 + 0.5 x 0 x 0.25²

   s = 0.45 m

  x position = 0.45 m

At t = 0.50 s

Considering horizontal motion of diver:-

Initial velocity, u =  1.8 m/s

Acceleration , a = -0 m/s²

Displacement, s = ?

Time, t = 0.50 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

   s = 1.8 x 0.50 + 0.5 x 0 x 0.50²

   s = 0.90 m

  x position = 0.90 m

At t = 0.75 s

Considering horizontal motion of diver:-

Initial velocity, u =  1.8 m/s

Acceleration , a = -0 m/s²

Displacement, s = ?

Time, t = 0.75 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

   s = 1.8 x 0.75 + 0.5 x 0 x 0.25²

   s = 1.35 m

  x position = 1.35 m

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pe = weightxheight = 2000x1000=2,000,000joules