Physics problem...I NEED HELP!!!?

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 28.5m/s and at an angle of 36.9 degrees above the horizontal. You can ignore air resistance.

At what two times is the baseball at a height of 9.50m above the point at which it left the bat?

Answer 1

Vertical component of its velocity,
v = (28.5) sin 36.9 degrees
= 17.11 m/s

For vertical motion,
y = vt - 4.9t^2
9.5 = 17.11t - 4.9t^2
4.9t^2 - 17.11t + 9.5 = 0
t = 0.69 s or 2.9 s

Answer 2

v0 = the vertical component of the velocity
t = time
g = acceleration of gravity = -9.8 m/s^2
s = vertical distance

s = v0t + (1/2)gt^2

so:
v0 = 28.5 tan(36.9)
v0 = 21.4 m/s

s = 21.4t - (1/2)9.8 t^2 = 9.5
4.9t^2 - 21.4t + 9.5 = 0

t = (21.4 +/- SQRT(457.96 - 4(4.9)(9.5))/(2*4.9)
t = 2.184 +/- SQRT(271.76)/9.8
t = 2.184 +/- 1.682

The times are: .502 seconds and 3.866 seconds

Check: Find the time of maximum height. This should be 2.184 seconds since the trajectory of the ball is symmetric about this point. And this point is reached when the vertical component of the velocity is 0 so:
v = 0 = v0 + gt
21.4 = 9.8t
t = 2.184 seconds

Also: s = v0t + (1/2)gt^2
s = 21.4(2.184) - 4.9(2.184)^2
s = 23.365 m is the max height of the ball.
The time it takes the ball to fall from this height to 9.5 m is:

s = (23.365 - 9.5) = (1/2)9.8t^2
13.865 = 4.9t^2
t = 1.682
So this would make the times the ball reaches 9.5 equal to the time at max altitude plus or minus this number and this agrees with the original calculation.
t = 2.184 +/- 1.682

Answer 3

.693s and 2.80s