Answer:
a
t = 1.235 s
b
[tex]h_{max} = 2.55 \ m[/tex]
c
[tex]v = -7.10 m/s[/tex]
Explanation:
From the question we are told that
The initial velocity is [tex]u = 5.0 \ m/s[/tex]
The take of height is [tex]h = -1.30 \ m[/tex]
The negative sign show that the height is on the negative y-axis when the take off point is consider as the origin
Generally from the kinematic equation we have that
[tex]s = ut + \frac{1}{2}at^2[/tex]
=> [tex]-1.30 = 5.0 t + \frac{1}{2} (-9.8)t^2[/tex]
Here g is negative given that the swimmers jump motion is against gravity
[tex]4.9t^2 - 5t - 1.30[/tex]
solving using quadratic formula we obtain that
t = 1.235 s
Generally the highest point is mathematically evaluated as
[tex]h_{max} = \frac{v^2 - u^2 }{2 * (-g) }[/tex]
Here v = 0 m/s since the velocity at the highest point is 0
[tex]h_{max} = \frac{0^2 - 5^2 }{2 * (-9.8) }[/tex]
[tex]h_{max} = 2.55 \ m[/tex]
Generally her velocity when the feet hits the water is mathematically evaluated from kinematic equation as
[tex]v = \pm \sqrt{ u^2 + 2 (-g) * (-h)}[/tex]
[tex]v = -7.10 m/s[/tex]
The negative value of the velocity is selected because the velocity is on the negative y-axis
