Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    It's a SUVAT question but I always get stuck on it. I have no idea how to find the time in the SUVAT formulas.

    A well is 50m deep. A stone is released from rest at the top of the well. Find how long the stone takes to reach the bottom of the well.
    Offline

    13
    ReputationRep:
    u=initial velocity=0ms^-1, a=acceleration=9.8ms^-2, s=displacement=50m.

    v^2=u^2+2as, use this to get v (final velocity at bottom of well).
    Then rearrange the formula v=u+at to find t, the time taken to fall to the bottom of the well.
    Offline

    2
    ReputationRep:
    Simpler way than above:

    s=ut+\frac{1}{2}at^2
    s=50m
    u=0ms-1
    a=g=9.8ms-2
    Find t.

    50=\frac{1}{2}\times9.8\times t^2

    Rearrange for t:

    t = \sqrt{\dfrac{50}{\frac{1}{2}\tim  es9.8}}

    Offline

    17
    ReputationRep:
    Or alternatively you could use s=ut+ \frac{1}{2} at^2, which eliminates the possibility of carrying over any mistakes made in calculating the velocity.
    Offline

    17
    ReputationRep:
    (Original post by maxfire)
    Simpler way than above:

    s=ut+\frac{1}{2}at^2
    s=50m
    u=0ms-1
    a=g=9.8ms-2
    Find t.

    50=\frac{1}{2}\times9.8\times t^2

    Rearrange for t:

    t = \sqrt{\dfrac{50}{\frac{1}{2}\tim  es9.8}}

    Beat me to it.
    Offline

    2
    ReputationRep:
    (Original post by Farhan.Hanif93)
    Beat me to it.
    To beat maxfire, you gotta be a fast mother ******.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by maxfire)

    50=\frac{1}{2}\times9.8\times t^2

    Rearrange for t:

    t = \sqrt{\dfrac{50}{\frac{1}{2}\tim  es9.8}}

    Thanks

    How do you know whether the gravity is - or + ?
    Offline

    2
    ReputationRep:
    (Original post by Mist786)
    Thanks

    How do you whether the gravity is - or + ?
    In this case, the ball is travelling downwards, in the same direction as gravity, so gravity is positive.

    If the ball was thrown upwards, gravity would be acting against it, so it would be negative
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by maxfire)
    In this case, the ball is travelling downwards, in the same direction as gravity, so gravity is positive.

    If the ball was thrown upwards, gravity would be acting against it, so it would be negative
    Oh, I get it

    Any ideas how this question is solved .. A particle is projected vertically upwards with speed 20ms-1 from a point on the ground. Find the time of flight of the particle.

    I used v=u+at but that gives me the wrong answer :rolleyes:
    Offline

    14
    ReputationRep:
    (Original post by Mist786)
    Oh, I get it

    Any ideas how this question is solved .. A particle is projected vertically upwards with speed 20ms-1 from a point on the ground. Find the time of flight of the particle.

    I used v=u+at but that gives me the wrong answer :rolleyes:
    I would use the displacement formula s=ut+\frac{at^2}{2} in this case the displacement is zero

    You could calculate the time for the velocity to decrease to zero. This would be when the ball reaches the top of its flight. To get time of flight you simply double.
    Offline

    2
    ReputationRep:
    (Original post by Mist786)
    Oh, I get it

    Any ideas how this question is solved .. A particle is projected vertically upwards with speed 20ms-1 from a point on the ground. Find the time of flight of the particle.

    I used v=u+at but that gives me the wrong answer :rolleyes:
    Yep, if you use
    s=ut+\frac{1}{2}at^2 again:

    s=0 (no overall displacement is moved when it comes down to the ground)
    u=20
    a=-g= -9.8 (it's going against the initial velocity).

    so:
    s=ut+\frac{1}{2}at^2

    0=20t+\frac{1}{2}\times-9.8t^2

    divide through by t:

    0=20-4.9t

    20=4.9t

    t=\frac{20}{4.9}=4.081 s

    Is that the right answer?

    It's all about knowing which equation to use tbh.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by maxfire)

    Is that the right answer?

    It's all about knowing which equation to use tbh.
    Yes, thanks again

    I always use the wrong equation, not always but most of the time.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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.