Sarah throws a ball to her friend Julie. What is the speed of the ball...?

When the ball reaches its maximum height, its vertical velocity is 0 m/s. The vertical acceleration is -9.8 m/s^2. Let’s use the following equation to determine the ball’s initial vertical velocity.

vf^2 = vi^2 + 2 * a * d, vf = 0, a = -9.8
0 = vi^2 + 2 * -9.8 * 8
vi = √156.8

This is approximately 12.5 m/s. Let’s use the following equation to determine the time for this to happen.

vf = vi – g * t
0 = √156.8 – 9.8 * t
t = √156.8 ÷ 9.8

This time is approximately 1.28 seconds. This is less than 1.44 seconds.
t = 1.44 – √156.8 ÷ 9.8

This time is approximately 0.162 second. Let’s use the following equation to determine the height of the ball at this time.

hf = hi + vi * t + ½ * g * t^2, hi = 8 meters, vi = 0 m/s, g = -9.8 m/s^2
hf = 8 + ½ * -9.8 * (1.44 – √156.8 ÷ 9.8)^2
This approximately 7.87 meters.

Edit
The ball reaches a height h = 6.5 m above from the throwing position, so its initial VERTICAL speed is sqrt(2gh) = 11.3 m/s using conservation of energy.

Horizontal velocity of the ball is of course unchanged throughout the flight, 23.057 m / 1.44 s = 16 m/s
So the total initial speed is sqrt(11.3^2 + 16^2) m/s = 19.59 m/s


b)
The height above ground of the ball as a function of time is
h = 11.3t - (9.8/2)t^2 + 1.5
Using the displacement equation for uniformly accelerated motion.
Where h is in m and t in s.
Plug in t=1.44 s.
h = 7.59 m