Respuesta :
Explanation:
Given that,
Velocity = 4.00 m/s
Distance = 1.80 m
(a). We need to calculate the time
Using equation of motion
[tex]s = ut+\dfrac{1}{2}gt^2+h[/tex]
Put the value in the equation
[tex]0=4.00t+\dfrac{1}{2}\times9.8\times t^2+1.80[/tex]
[tex]-9.8t^2+8.00t+3.60=0[/tex]
[tex]t =1.139\ sec[/tex]
The time is 1.139 sec.
(b). We need to calculate the highest point above the board her feet reach
Using equation of motion
[tex]v^2-u^2=2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2g}[/tex]
Put the value in the equation
[tex]s=\dfrac{0^2-(4.00)^2}{2\times(-9.8)}[/tex]
[tex]s=0.82\ m[/tex]
The highest point above the board her feet reach is
[tex]s' = 0.82+1.8 = 2.62\ m[/tex]
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
[tex]v = u+gt[/tex]
Put the value in the equation
[tex]v=4.00-9.8\times1.139[/tex]
[tex]v=-7.16\ m/s[/tex]
Her velocity when her feet hit the water is 7.16 in downward.
Hence, This is the required solution.
