Basketball physics in two-dimensions?

Question

The question I am given is: "A basketball star covers 2.60 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled as that of a particle at a point called his center of mass (which we shall define in Chapter 9). His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.80 m above the floor and is at elevation 0.850 m when he touches down again."

And i need to find the time of flight, the initial horizontal/vertical velocity, and angle.

However, I am pretty clueless how to find the total time. I figured x(final)=.5at^2 So sqrt((0.95*2)/a)=time, but 0.44, the answer i get (times two, so 0.88) isn't the overall hang time. Now i'm just completely lost.

Answer 1

First you need to figure out the flight time.

Rise distance: drise = peak cm height - start cm height
drise = 1/2gt^2
So rise time = sqrt (2 drise / g)
Fall distance: dfall = peak cm height - end cm height
So fall time = sqrt (2 dfall / g)
Total flight time = sqrt (2 (peak height - start height)/g)
+ sqrt (2 (peak height - finish height)/g)

You can then calculate horizontal velocity
vh = distance covered / total flight time

You can calculate initial vertical velocity using conservation of energy on the vertical motion

initial vertical KE = final PE
1/2 m vyi^2 = mg drise
vyi = sqrt (2 g drise) = sqrt (2 g (peak height - start height))

To get the jump angle
takeoff elevation = arctangent (vyi / vx)

Answer 2

He rises ymax=0.95 m from starting position and travels x=2.6 m horizontally. His initial velocity v0 can be defined as follows (using g=9.8 m/s^2):
v0y = sqrt(2g(ymax-y0)) = 3.910 m/s
t1 = sqrt(2(ymax-y0)/g) = 0.399 s
t2 = sqrt(2(ymax-y1)/g) = 0.440 s
t = t1+t2 = 0.839 s
v0x = x/t = 3.098 m/s
v0 = sqrt(v0x^2+v0y^2) = 4.988 m/s
theta = arctan(v0y/v0x) = 51.61 deg
EDIT: Whoops. Didn't read carefully; answer above changed to accommodate different initial and final heights y0 and y1.