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?

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.

Answered by taskmasters | 2024-06-10

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:
S = 2 1 ​ g t 2 = 2 1 ​ ( 9.81 m / s 2 ) ( 1.1 s ) 2 = 5.93 m

Answered by skyluke89 | 2024-06-11

The maximum height of the volleyball serve is approximately 5.94 meters, which is calculated using the time the ball was in the air and the acceleration due to gravity. Since the ball takes 2.2 seconds total, it reaches its peak at 1.1 seconds. The kinematic equation for displacement under constant acceleration is used for the calculation.
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Answered by skyluke89 | 2024-09-30