A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 19.6t + 58.8, where s is in meters. When does the object strike the ground @zepdrix

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

    \[\large\rm s(t) = –4.9t^2 + 19.6t + 58.8\]Mmm k so this is the same as the third part of the last problem. We care about the time, t, at which the height of the function, s(t), is zero. So we'll let s(t) be zero and then plug all the goodies into the quadratic formula.\[\large\rm 0 = –4.9t^2 + 19.6t + 58.8\]Plug em in! :O\[\rm t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

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

    |dw:1441258743147:dw|

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

    What's under the square root? :O It should be 1500 or something

  4. 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.