Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

a batting machine launches a ball straight up in the air at a velocity of 100 ft/s. the machine is 65 feet tall what is the maximum height the ball can attain how fast will it be moving at this point?

I got my questions answered at in under 10 minutes. Go to now for free help!
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


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

Well, let's start from the top. It's important to be able to derive the equations you need for physics without relying on memorization. Do you know calculus?
Good. Assume that the ball, at any given moment, has constant acceleration \(a\). First, find a formula for \(v(t)\), the vertical velocity of the ball at any given time.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I know that it needs to involve a (which would be negative due to gravity) and x over a certain period of time t. So \[v(t)=a(\int\limits_{0}^{t}xdt)\]?
Well don't worry about what the value of \(a\) yet (although you are right in that it is negative and the gravitational acceleration constant, which is \(-g = -9.8 \text{m}/\text{s}^2\). Solving for \(v(t)\) should be easy. If the derivative (acceleration) is constant, then \(v(t) = a t + C\) right?
Ugh! Yes. It has been so long since I have done any word problems like this and it frustrates me that I can't remember how to start them off. I am really trying to help my cousin with her homework :/
Ok. well, we still have on more time-integral left. since this isn't for you, I'll just give you the solution: \(y = y_0 + v_0 t + a t^2/2\) I trust you can find approp. values for the constants, right?
\[y_{0}=65 ft v _{0}=100ft/s \]
Thank you so much for the help. I am clearly too tired after my own exams to be helping anyone else.
Yes @ your values. Max height can be found using single-derivative optimization (find where \(v(t) = 0\)).

Not the answer you are looking for?

Search for more explanations.

Ask your own question