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    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.
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    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.
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    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}}

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    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.
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    (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.
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    (Original post by Farhan.Hanif93)
    Beat me to it.
    To beat maxfire, you gotta be a fast mother ******.
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    (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 + ?
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    (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
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    (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:
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    (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.
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    (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.
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    (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.
 
 
 
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