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Kinematic Equation Questions - Ball Verticle Height

  1. Mar 27, 2008 #1
    1. The problem statement, all variables and given/known data
    A bell is thrown vertically upwards with speed 10m/s from a point 2 m above horizontal ground.
    a) calculate the length of time for which the ball is 3m or more above the ground

    2. Relevant equations

    kinematic ones

    3. The attempt at a solution

    We I wanted to find the intial speed of the ball when it is 3 metres above the ground

    i used v² = u² + 2as

    and found u to be about 10.936

    now do i need to find how high the ball goes from 3m above the ground to having a velocity of zero? I got 314/93. Once i find this i can I then use this in the s= ut + 0.5at² equation.

    4.9t² + 10t - (314/53) = 0

    I tried and i got a final answer of -1.86 when the answer in the book is 1.83

    Have I done the right steps???

    Last edited: Mar 27, 2008
  2. jcsd
  3. Mar 27, 2008 #2
    The initial speed IS 10 m/s. A way to approach the problem is to have the displacement above ground relative to your initial displacement.
  4. Mar 27, 2008 #3
    this intial speed of the ball at 2m is 10 but at 3 m it's 10.936 i thinnk

    i used v² = u² + 2as

    v = sqrt(100+2(9.8)(1))
    = 10.936
    is that the right thing to do?
  5. Mar 28, 2008 #4
    never mind. i managed to solve part a in the end. I'm now stuck on part b

    b) Calculate the speed with which the ball hits the ground.

    So first of all calculate the parabola bit from 2 metres to it's greatest height then back to 2m so the overall displacement (s) is 0

    use s = ut + 0.5at²
    0 = 10t + 0.5(9.8)t²
    so t = 0 or t = -(100/49)

    then work out the last bit of the journey below the 2meters?

    I'm not sure what to do. Can someone guide me through it step by step

    the answer is 11.8s in the answer booklet

  6. Mar 28, 2008 #5


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    Things are easier if you keep your directions consistent. u is 10m/sec UPWARDS. g is 9.8m/sec^2 DOWNWARDS. So I would write s=t*10m/sec+(1/2)*(-9.8m/sec^2)t^2. Now if s=0 is 2m above the ground, the ground is located at s=(-2). Find t at s=(-2).
  7. Mar 28, 2008 #6


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    The only difference between t0, when the ball was headed upward, and it's speed when it passed the 2m mark in the downward direction is the sign. Your initial speed is -10 m/s instead of +10 m/s. You know your final position is 0. Calculate the time it takes to go from 2m to 0, then solve for your final velocity using v=u + at
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