Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

tini12

  • one year ago

Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 20 degrees. It reaches the maximum height of 15 m. The acceleration due to gravity is 15 m/s^2. Find the minimum speed of the projectile during the flight. Enter the answer in "m/s".

  • This Question is Closed
  1. tini12
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You are participating in an unusual drag race. You start from rest and try to cover a straight quarter-mile (400 meters, for the purpose of this problem) in the shortest time possible. Your car is very well built so it can achieve speeds of up to 360 km/h. It is equipped with a very powerful engine and an excellent anti-lock braking system. The coefficient of static friction between the wheels and the road is about 1.2 (the drag-race tires are made of very soft rubber which is pre-heated before the start by spinning the wheels; as a result, the wheels achieve the effective coefficient of friction greater than 1). However, there is a crucial difference: you are required, for publicity purposes, to come to a dead stop at the finish line (in a typical drag race, you would fly right through that finish line!). speed up at the maximum possible rate for the first half of the distance; then slow down at the same rate What is the shortest possible time needed to complete the race (in seconds)?

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

    For the first problem the flat tundra thing. Recall projectile motion. At it's peak, there is horizontal motion. At any other point, there is vertical velocity. Since there is no horizontal acceleration, and by bringing back newton's first law, the minimum speed is where the vertical speed is zero, and thus finding the horizontal speed would suffice. Now the problem to find the horizontal distance. Recall that: \[v^2=u^2+2as\] So we find that u=21.2m/s Resolving, \[V_{vertical}=V_{0}Sin 20^{o}=21.2m/s\]and \[V_{horizontal}=V_{minimum}=V_{0}Cos20^{o}\] Solving for , v minimum is then 58.3m/s

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

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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.