Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

skeller2

  • 2 years ago

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?

  • This Question is Open
  1. vf321
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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?

  2. skeller2
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes

  3. vf321
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  4. skeller2
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \[v(t)=a(\int\limits_{0}^{t}xdt)\]?

  5. vf321
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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?

  6. skeller2
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 :/

  7. vf321
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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?

  8. skeller2
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[y_{0}=65 ft v _{0}=100ft/s \]

  9. skeller2
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you so much for the help. I am clearly too tired after my own exams to be helping anyone else.

  10. vf321
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes @ your values. Max height can be found using single-derivative optimization (find where \(v(t) = 0\)).

  11. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.