anonymous one year ago A volleyball is thrown up in the air with initial velocity of 8.2m/s. what is the maximum height reached?

1. zephyr141

use the kinematic equations. looks like it's the same ball with the same initial velocity. so that means we can use the time we found in the last question to find the maximum height achieved. now remember that at it's highest point the final velocity in the y direction will be 0 so knowing that let's look at the kinematic equations again and pick the best equation that best fits this problem.

2. zephyr141

matter in fact we can use this one$d=v_0t+\frac{1}{2}at^2$

3. zephyr141

i think the time we found was 0.84 seconds.

4. anonymous

well my teacher said not to use the same time for this cause it'll give me the wrong answer

5. zephyr141

are you sure? because the time isn't going to change if you solve for time for this equation. i just checked and the velocities are the same in both questions and since gravity isn't changing it has to be the same time.

6. anonymous

i didnt understand her reasoning but i think thats whats throwing me off.

7. zephyr141

ok lets leave time out then. use this equation. it doesn't need time.$v^2=v_0^2+2ad$

8. zephyr141

isolate d in the equation first.

9. anonymous

alright

10. zephyr141

you should get $d=\frac{v^2-v_0^2}{2a}$and now just plug in your known values. v0=8.2 m/s v=0 a=-9.8

11. anonymous

okay. so i got -8.2/ -19.6??

12. zephyr141

$d=\frac{0-(8.2)^2}{-19.6}=\frac{-67.24}{-19.6}=3.4306m$ or 3.4 meters. which is the same if we used time. so i don't really know why your teacher told you not to use time. maybe to solve it using a different equation.

13. anonymous

oh man i didnt square them. yah probably idk ill ask her about it, thanks! again lol

14. zephyr141

lol no problem.