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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?

Sagot :

Taskmastersidn
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.



Skyluke89idn

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]

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