PHYSICS!! Please Help..(how high, how long?)?

Question

A volleyball is served from a heigh of 0.78m and gives it an initial velocity of +7.7 m/s straight up.
a.) How high will the volleyball go?(m)
b.) How long will it take the ball to reach its maximum height?(s)

Answer 1

First calculate how long it takes the ball to reach maximum height and then use this to calculate the height.

v = v0 + gt

At max height v = 0 so v0 = -gt and t = -v0/g

v0 = 7.7 m/s
g = -9.8 m/s^2
t = -v0/g = 7.7/9.8 = 0.786 seconds

a.
s = v0t + (1/2)gt^2
s = 7.7(0.786) - 4.9(0.786)^2
s = 3.025 meters

You can also use v^2 = v0^2 + 2gs or s = -v0^2/2g
s = (7.7)^2/(2*9.8)) = 3.025

Since the ball is at 0.78 meters above the ground when it is served, this height must be added to "s" to get the maximum altitude reached by the ball.

Height = 0.78 + 3.025 = 3.805 meters

b. This was worked out at the start and is 0.786 seconds

Showing v^2 = v0^2 + 2gs
I always remember v = v0 + gt and s = v0t + (1/2)gt^2
From the first: t = (v - v0)/g
Put in the second: s = v0(v - v0)/g + (1/2)g[(v - v0)/g]^2
2sg = 2(v0v - v0^2) + (v- v0)^2
2sg = 2v0v - 2v0^2 + v^2 - 2v0v + v0^2
2sg = v^2 - v0^2
v^2 = v0^2 + 2sg
I now remember this but I used to not remember and it is always nice to know how something can be derived.