A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

a man in a balloon rising vertically with an acceleration 4.9ms release a ball 2 seconds after the balloon is let go from the ground the maximum height above the ground reached by the ball is

  • This Question is Open
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if the ball is released, its initial speed is 0, and since afterwards it goes down due to gravity, the max height of the ball is the height at which is dropped. so to get the answer you need to calculate the height of the ballon at t=2 seconds. it moves with constant accelertion so you can use: \[y(t)=y_0+v_0t+\frac{1}{2}at^2\] evaluate for t=2, and think what the values for initial height (y0) and initial speed (v0) should be for the balloon

  2. BAdhi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Greg_D At the time of the release of the ball dont you think it will have the velocity of the balloon? So at the height that is calculated from your equation \(y(t)\) it will have a velocity upwards. So before starting to fall the ball will move upwards for a while after it is released. I think that extra distance should be added as well

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes! you are right @BAdhi ! is not as i said before, we need first to calculate the speed of the balloon when the ball is released, that will be the initial speed of the ball : \[v_{balloon}(t) = at\] evaluate that for t=2, and evaluate y(t) for the balloon at t=2 to get the height. then for the ball we have: \[y_{ball}(t)=y_0+v_0t-\frac{1}{2}gt^2\] with y0, the height of the balloon at t=2 and v0 the speed of the balloon at t=2. thanks for the correction @BAdhi !

  4. BAdhi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \(y_{ball}(t)\) can easily be taken by \(v^2= u^2+2as\), where \(s=y_{ball}\) and \(u=v_{ballon}\). since we do not know the time taken for the ball to reach its maximum height

  5. 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.