Question

Projectile question:
A basketball player jumped straight up to grab a rebound. If she was in the air for 0.80 seconds, how high did she jump?

Answer: She was in the air for 0.80 seconds, so this means that it took her 0.4 seconds to reach her maximum height and an additional 0.40 seconds to fall back to the ground. Therefore we
have the freefall equation: x =1/2(9.8 m/s^2)(0.40 s)^2 = 0.784 m.
Question: Why is the initial velocity 0 m/s^2? She must have had a velocity to jump up.

Answer 1

The answer is correct when it splits the total time in half, as the time up and the time down should be equal.

We could look at this problem from the falling perspective first. Consider that the player just caught the ball and is going to fall (initial velocity is 0 m/s since the player is at the max height). Then the player would fall according to the equation the answer gives, which gives you a height of 0.784 m.

I think what is confusing you (and is kind of incompletely described) is the coordinate system used. You have to choose a zero point. The answer obviously uses the zero point as the spot where the player is at her maximum height.
You are correct. The player needs an initial velocity to propel herself upwards to get the ball. And we can actually find it. We just need to stay true to our coordinate system.

So we have the kinematic equation:
X = Xo + Vo * t + 0.5 * a * t^2

Now, using our coordinate system:
Xo = 0.784 m
X = 0 m
Vo = ? m/s
t = 0.4 s
a = g = 9.8 m/s^2

This gives us:
0 = 0.784 - 0.4*Vo + 0.5*9.8*(0.4^2)
-0.784 = -0.4*Vo + 0.784
1.586 = 0.4*Vo

Vo = 3.92 m/s

So as you can see, the initial velocity is indeed not zero! You were correct. It was simply the unexplained coordinate system. X = 0 was at the top of the jump.