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Cricket ball / projectile question

  1. Nov 14, 2006 #1
    I am trying to repeat a quesion I had got wrong about a cricket ball. The question is

    "A cricketer bowls a ball from a height of 2.3m. The ball leaves his hand horizontally with a velocity u. After bouncing once, it just passes over the stumps at the top of its bounce. The stumps are 0.71m high and are situated 20m from where the bowler releases the ball.
    a) show that from the moment it is released the ball takes about 0.7s to fall 2.3m
    b) How long does it take the ball to rise 0.71m after bouncing?"

    Part a I'm ok with. I used v^2 = u^2 +2ax using
    a = -9.81m/s2
    x = 2.3m
    u = 0m/s
    to find v = -6.72 m/s

    i then used v = u +at
    to give t = (v-u)/a = 0.68s

    For part b I thought I could use a similar approach saying that at the top of it's bounce the vertical component of the velocity will be zero and the only acceleration is due to gravity. So using v^2 = u^2 +2ax
    with v= 0m/s
    a = -9.81m/s2
    x = 0.71
    I got u = 3.73 m/s

    and substituing that into t =(v-u)/a, t = 0.38s
    unfortunately this was wrong and I got 0 out of a possible 3 marks.

    Please can anyone help me with where I am going wrong? The only thing I can think of is that there is some additional acceleration in the vertical direction (from the bounce??) but I don't know how to work this out. If you can point me in the right direction that would be brilliant.
     
  2. jcsd
  3. Nov 14, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I don't see anything wrong with your work. Your answers look correct to me.
     
  4. Nov 15, 2006 #3
    Thank you!
     
  5. Nov 15, 2006 #4
    Hey I’m not too sure, but wouldn’t the x value in the second vertical part be larger as it travels over the wickets, it says that it is at its max height over the wickets, so all that that is saying is that its max bounce height is 20 meters from where the ball was released
     
  6. Nov 15, 2006 #5
    oh wait sorry im wrong
     
  7. Nov 15, 2006 #6
    no wait im sorta right! lol yea u can't sub V=0 when x=0.71, because the ball travels over the wickets!
     
  8. Nov 15, 2006 #7
    lol so X > 0.71. when V= 0.
     
  9. Nov 15, 2006 #8
    I says it 'just' passes over the stumps at the top of its bounce so doesn't that mean the top of the bounce is at 0.71m?
     
  10. Nov 15, 2006 #9
    yea ok maybe, lol better ask your teacher, lol i kno i would have not got that question right either! id say you would have to include 20m some how because otherwise it wouldnt have been given. lol good luck lol
    lol and let me kno please!
     
  11. Feb 24, 2007 #10
    yeh...you have to include that there is 20 m that they ball has to travel because 0.38 s for a ball to travel 20 m is far to fast...i still think it is pretty harsh that you didnt get any marks...
     
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