Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    A small piece of dung is fired into the air at 30ms^-1. How long does it take to reach a height of 5m?

    i know I need to use suvat equation do im using x=ut+1/2 at^2

    (x)= displacment.

    I presume i need to turn this into a quadratic of t in order to solve it but im not to sure how to go about this anyone give a hand.
    Offline

    10
    What is it you don't know?
    How to turn it into a quadratic, or how to solve the quadratic?
    1) Sub the values you know into the formula and move x to the other side to get at² + bt + c=0
    a b and c from the values given in the question and the acceleration due to gravity.
    Then use the formula for solving a quadratic. [Do you know this?]
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Stonebridge)
    What is it you don't know?
    How to turn it into a quadratic, or how to solve the quadratic?
    1) Sub the values you know into the formula and move x to the other side to get at² + bt + c=0
    a b and c from the values given in the question and the acceleration due to gravity.
    Then use the formula for solving a quadratic. [Do you know this?]
    Ah sorry didn't make myself clear. Ye i just wasnt sure how to turn it into a quadratic. So to do this I use at^2=a ut=b and x =c ? If so what happens to the half?
    Offline

    10
    Double post. The better answer is in the other one!
    Offline

    2
    ReputationRep:
    (Original post by hazbaz)
    Ah sorry didn't make myself clear. Ye i just wasnt sure how to turn it into a quadratic. So to do this I use at^2=a ut=b and x =c ? If so what happens to the half?
    x=ut+1/2at²

    rearranges to:

    1/2at²+ut-x=0

    Get rid of the 1/2 by multiplying everything by 2:

    at²+2ut-2x=0

    (Where x is displacement, u is initial velocity, a is acceleration [deceleration] due to gravity, t is time taken to travel x distance.)



    Stonebridge's point in using at²+bt+c=0 (i think) was so you could relate it to ax²+bx+c=0, where you can use x=-b +/- root (b²-4ac), all over 2a.

    Sorry for the lack of latex this time.

    So, a will remain acceleration, b would be 2u and c would be -2x (where x is displacement)
    Offline

    2
    ReputationRep:
    Double post. Didn't notice second topic lol.
    GJ.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Ye thanks i got it
 
 
 
The home of Results and Clearing

2,246

people online now

1,567,000

students helped last year

University open days

  1. Sheffield Hallam University
    City Campus Undergraduate
    Tue, 21 Aug '18
  2. Bournemouth University
    Clearing Open Day Undergraduate
    Wed, 22 Aug '18
  3. University of Buckingham
    Postgraduate Open Evening Postgraduate
    Thu, 23 Aug '18
Poll
How are you feeling about GCSE results day?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.