A volleyball serve was in the air for 2.2 seconds before it landed untouched in the far corner of the 
opponent’s court. What was the maximum height of the serve?

We are only interested in the vertical motion of the ball.The ball remains in air for t=2.2 s, so we can say that it reaches its maximum height in t=1.1 s (half the time) before falling down. This is an uniformly accelerated motion with constant acceleration g=9.81 m/s^2, so the maximum height reached by the balls is given by:[tex]S=\frac{1}{2}gt^2 = \frac{1}{2}(9.81 m/s^2)(1.1s)^2=5.93 m[/tex]

Given that
air resistance is negative.

Hence the
values are:

2.2 s

G = 9.8
m/s2

Solution

Velocity =
vf-vi = 0 – vi = -vi

V = 9.8m/s2
x 2.2 s = 21.56 m/ s (cancel one s out)

Vi = 21.56
m/s

 

Then (s) displacement
is

S=v x t

Average velocity
= (21.56 m/s + 0 / 2)

Average
velocity = 10.78 m/s

Time = 2.2
s

 
S = 10.78
m/s x 2.2 s

(cancel s)S = 23.716
m

 

Therefore
the ball was at 23.716 meters in the air.