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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Physics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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.
matter in fact we can use this one\[d=v_0t+\frac{1}{2}at^2\]
i think the time we found was 0.84 seconds.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

well my teacher said not to use the same time for this cause it'll give me the wrong answer
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.
i didnt understand her reasoning but i think thats whats throwing me off.
ok lets leave time out then. use this equation. it doesn't need time.\[v^2=v_0^2+2ad\]
isolate d in the equation first.
alright
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
okay. so i got -8.2/ -19.6??
\[d=\frac{0-(8.2)^2}{-19.6}=\frac{-67.24}{-19.6}=3.4306m\] 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.
oh man i didnt square them. yah probably idk ill ask her about it, thanks! again lol
lol no problem.

Not the answer you are looking for?

Search for more explanations.

Ask your own question