Answered

Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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 :

taskmasters
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.



skyluke89

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]

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.