Abasketball player jumps straight upward. after 0.625 s, she is 0.411 m above the ground. what was her initial velocity?
use the formula
trianglex = vi * t+ 1/2g * t^2

Formula for the height is:
h = v i · t - 1/2 g · t²    ( minus sign because the gravitational force is in the opposite direction then the force that a player exerts )
0.411 = v i · 0.625 - 1/2 · 9.81 · 0.625²
0.411 = 0.625 v i - 1.916
0.625 v i = 0.411 + 1.916
0.625 v i = 2.327
v i = 2.327 : 0.625 = 3.72 m /s
Her initial velocity was 3.72 m/s.

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