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Throws a stone horizontally with a velocity of 20ms-1

  1. Sep 12, 2012 #1
    1. The problem statement, all variables and given/known data

    1. Don Joriel stands on a vertical cliff edge throwing stones into the sea below. He throws a stone horizontally with a velocity of 20ms-1, 560 m above sea level.
    a. How long does it take for the stone to hit the water from leaving Don Joriel’s hand?
    Use g = 9.81 ms-2 and ignore air resistance?
    b. Find the distance of the stone from the base of the cliff when it hits the water.

    2. When Maggie applies the brakes of her car, the car slows uniformly from 15.00 m/s to 0.00 m/s in 2.50 s. How many meters before a stop sign must she apply her brakes?



    2. Relevant equations



    3. The attempt at a solution
    please help me with those problem, im confused. I am not sure if i got the correct answer. here are my answers 1. a. 2.04 s
    b. 20.39 m
    2. 18.75 m

    Hope you can help me, if got the wrong answer please show me the solution... Thanks a lot...
     
  2. jcsd
  3. Sep 12, 2012 #2

    rl.bhat

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    Re: motion

    Show your calculations.
     
  4. Sep 12, 2012 #3

    Simon Bridge

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    Re: motion

    What rl.bhat said: if you don't show us what you tried or your reasoning, then we cannot know how to help you best.
     
  5. Sep 12, 2012 #4
    Re: motion

    wait i'll show you when i got home.. thanks...
     
  6. Sep 12, 2012 #5
    Re: motion

    for 1a. i use v = u + at,
    b. s = ut +1/2 at squared

    for 2. s = 1/2 (u +v)t

    are those correct?
     
  7. Sep 12, 2012 #6

    CAF123

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    Re: motion

    I don't see how you can use the equation in 1a without first knowing the final velocity of the ball (I.e the velocity of the ball prior to impact with the ground). From your answer, it seems you have assumed this to be 0, but this is not the final vertical component of velocity.
     
  8. Sep 12, 2012 #7

    CAF123

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    Re: motion

    For 1b, we are considering a horizontal distance. There is no acceleration of the ball in x direction so the eqn you were using simplifies to simply [itex] s = v_{ox}t [/itex].
    For 2, if I understand the wording of the question correctly, you are correct.
     
  9. Sep 12, 2012 #8
    Re: motion

    what is the answer then? please show the solution...
     
  10. Sep 12, 2012 #9

    rl.bhat

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    Re: motion

    Before you attempt this problem, please go through the projectile motion.
     
  11. Sep 12, 2012 #10

    CAF123

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    Re: motion

    We are not allowed to simply hand over solutions. For 1a, you can use the equation [itex] s_y = v_{oy}t - \frac{1}{2}gt^2 [/itex] right away to find the time of descent. Alternatively, the final vertical component of velocity can be found first so you can then use [itex] v_{fy} = v_{oy} -gt [/itex].
    Yet another way is to do all three problems via sketching v-t diagrams for each of the x and y components of velocity.
     
  12. Sep 12, 2012 #11
    Re: motion

    what is the answer on the problem? i am totally confused.. please help me...
     
  13. Sep 12, 2012 #12
    Re: motion

    i didnt get your equation because of so much symbols, sorry...
     
  14. Sep 12, 2012 #13
    Re: motion

    i got it already... thanks...
     
  15. Sep 12, 2012 #14
    Re: motion

    how about number 2?
     
  16. Sep 12, 2012 #15

    rl.bhat

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    Re: motion

    Can you find the time taken by a stone dropped from a height 560 m?
     
  17. Sep 12, 2012 #16

    CAF123

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    Re: motion

    Ok, for 1a, initial vertical component of velocity is 0, right? This is because the ball is projected horizontally. So in our equation, [itex] v_{oy} =0. [/itex] The eqn therefore reduces to [itex] s_y = \frac{1}{2}at^2. [/itex] If you define downwards as negative, then the ball will fall downwards 500m so [itex] s_y = -500m. [/itex] Think of the ball moving 500 units down the y axis (I.e starts at y =0 and finishes at y = -500). Now, if downwards is negative, a =-g and so our final equation is [itex] -s_y = -\frac{1}{2}gt^2 [/itex] Rearrange for t and you are done with 1a.
     
  18. Sep 12, 2012 #17
    Re: motion

    so t = 10.68... by the way where did you get 500? is it 560?
     
  19. Sep 12, 2012 #18
    Re: motion

    how about 1b?
     
  20. Sep 12, 2012 #19

    CAF123

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    Re: motion

    Yes, sorry 560m not 500m. For 1b, again use the relation [itex] s_x = v_{ox}t + \frac{1}{2}at^2, [/itex] but this time considering the x direction. No forces act on the body in the horizontal direction so ax = 0. The horizontal distance covered will be proportional to the time of flight because of a constant (non zero) horizontal component of velocity.
     
  21. Sep 12, 2012 #20

    Simon Bridge

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    Re: motion

    Please tell us what you got... be specific: it makes sure we have not accidentally sent you up the wrong tree. When you show working, also be specific: we need to see your reasoning, like how you used an equation, in order to help you properly.

    CAF is doing the detail work. Can you see how CAF is choosing which equation to use?
     
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