Answered

Explore a world of knowledge and get your questions answered on IDNLearn.com. Discover in-depth answers from knowledgeable professionals, providing you with the information you need.

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]

Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is dedicated to providing accurate answers. Thank you for visiting, and see you next time for more solutions.