The time to rise to maximum height will be the same time that it takes to fall.
Take your 1.45 seconds, and divide by 2. This gives you 0.725 seconds. This is the time that it takes to rise to its maximum height. It is *also* the time that it takes to fall back down to the baseball player.
You can use this equation which figures out time to maximum height:
Tf = SQRT [ (2H) / g ]
Where Tf is the time to rise (0.725 seconds), H is the maximum height, and g is Earth gravity (9.80665 m/s^2).
Isolate the H
Tf^2 = (2H) / g
(Tf^2) * g = 2H
[ (Tf^2) * g ] / 2 = H
Now just plug in your known time Tf and gravity
H = [ (0.725 s)^2 * (9.80665 m/s^2) ] / 2
H = [ (0.525625 s^2) * (9.80665 m/s^2) ] / 2
H = 5.155 m
Note that the units cancel out.
This is the answer to "Maximum Height"
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5.155 m
Initial velocity is going to be the same as *final* velocity, because that's the way it works. So go to this equation:
Tr = (Voy) / g
Where Tr = time to rise (which is the same as time to fall, as above! 0.725 s) and Voy is initial velocity.
Solve for Voy
Tf * g = Voy
Insert values
Voy= (0.725 s) * (9.80665 m/s^2)
Voy = 7.110 m/s
Which is your answer to Initial Velocity
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7.110 m/s
It's true that you can do it in a single equation H = Yo + (Voy * Tr) - (0.5 * g * Tr^2) ... but I prefer breaking it into simple pieces.