Question

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) How long are her feet in the air?� (b) What is the highest point above the board her feet reach? (c) What is her velocity when her feet hit the water?

Answer 1

Step-by-step explanation:

Given that,

Velocity = 4.00 m/s

Distance = 1.80 m

(a). We need to calculate the time

Using equation of motion

`s = ut+(1)/(2)gt^2+h`

Put the value in the equation

`0=4.00t+(1)/(2)*9.8* t^2+1.80`

`-9.8t^2+8.00t+3.60=0`

`t =1.139\ sec`

The time is 1.139 sec.

(b). We need to calculate the highest point above the board her feet reach

Using equation of motion

`v^2-u^2=2as`

`s=(v^2-u^2)/(2g)`

Put the value in the equation

`s=(0^2-(4.00)^2)/(2*(-9.8))`

`s=0.82\ m`

The highest point above the board her feet reach is

`s' = 0.82+1.8 = 2.62\ m`

The highest point above the board her feet reach is 2.62 m

(c). We need to calculate her velocity when her feet hit the water

Using equation of motion

`v = u+gt`

Put the value in the equation

`v=4.00-9.8*1.139`

`v=-7.16\ m/s`

Her velocity when her feet hit the water is 7.16 in downward.

Hence, This is the required solution.