Question

A 754 N diver drops from a board 9.80 m above the water’s surface. The acceleration of gravity is 9.81 m/s 2 . a) Find the diver’s speed 4.70 m above the water’s surface. Answer in units of m/s.

Answer 1

`v=10.0031\ m.s^(-1)`

Step-by-step explanation:

Given:

• weight of the diver,
  `w=754\ N`

• height of descend,
  `h'=9.8\ m`

• height of observation,
  `h=4.7\ m`

Displacement of the diver:

`s=h'-h`

`s=9.8-4.7`

`s=5.1\ m`

Now using the equation of motion:

`v^2=u^2+2g.s`

where;

`v=` final velocity of the diver at the observed time

`u=` initial velocity of the diver when at the top height = 0

`v^2=0^2+2* 9.81* 5.1`

`v=10.0031\ m.s^(-1)`