Question

A diver 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.

Answer 1

Thank you for posting your question your question here. Below is the solution:
time = 3.00 / 1.75 

time = 1.714 

now 

distance = (initial velocity + final velocity) / 2 * time/ 1 

Hence

3 = ( 1.75 + final velocity) / 2 * 1.714 

3.5 - 1.75 = final velocity 

Therefore, the final velocity is 1.75 m /s

Answer 2

Answer:

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