A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 5 years ago

A balloon released at point A rises vertically with a constant speed of 5m/s. Point B is level with and 100m distant from point A. How fast is the angle of elevation of the balloon at B changing when the balloon is 200m above A?

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    let at any time t height of balloon is x m. and angle of elevation is y...we have to find dy/dt. tan y=x/100 (sec^2y)dy/dx=1/100 dy/dx=1/(100sec^2y) (dy/dt)(dt/dx)=1/(100sec^2y) dy/dt=1/(100sec^2y) .(dx/dt) put the value from question dy/dt=1/(100sec^2y).(5)....................1 now sec^2y=1+tan^2y=1+(200/100)^2=5 put this value in equation 1 dy/dt=1/500.(5) dy/dt=1/100

  2. 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...


  • 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.