Please help with my problem and show work as an example

Question

A diver dives off the 10-meter platform into the swimming pool below. If her initial velocity is 5.5 m/s, what is her maximum height above the platform, and when will she hit the water?

Answer 1

The maximum height is attained when the velocity is zero.

 

v_f² = v_i² + 2aΔy

 

0 = 5.5² + 2(-9.8)Δy

 

Δy = 1.54 m

 

 

She will hit the water when Δy = -10 m.

 

Δy = v_it + ½at²

 

-10 = 5.5t - 4.9t²

 

4.9t² - 5.5t - 10 = 0

 

t = 5.5/(2*4.9) +(1/(2*4.9))√(5.5² + 4(4.9)(10)) = 2.10 s