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How high is the bridge above the water?

  1. Jan 11, 2006 #1
    No 16 page 96: How long does it take for a car accelerating at 3.5 m/s^2 to go from rest to 120 km/h . Hint: first, change the velocity from km/h to m/s.





    No. 17 pag. 105: A rock is dropped from a bridge to the water below. It takes 2.4s to hit the water. find:



    a) the speed in m/s of the rock when it hits the water ( assume g,the acceleration and v , positive going down)



    b) how high is the bridge above the water??


    How would i solve this????Thanks in advanced
     
  2. jcsd
  3. Jan 11, 2006 #2
    For the second question (2b, more specifically), you can use the average speed. Now ask yourself, what do you need to calculate the average speed?

    I hope you do understand 2a.

    For the first question. You have the amount of speed increasement per second (definition of acceleration, roughly), and you have the final speed.

    If you need any more hints (it wasn't quite clear what you did and did not understand), reply.
     
  4. Jan 11, 2006 #3
    well for the second equation i dont know wht equation to use i have only time and acceleration of gravity which is 9.8
     
  5. Jan 11, 2006 #4
    i just need the equation
     
  6. Jan 11, 2006 #5
    i aleady solved the first one and got T=9.6s
     
  7. Jan 11, 2006 #6
    to calculate aveg speed i need vf-vi/t
     
  8. Jan 11, 2006 #7
    but i dont have vf or vi
     
  9. Jan 11, 2006 #8
    ahm, average spped would be (vf-vi)/2. 2a is quite easy, you have the amount of speed added per second, you know the number of seconds, so...
    2b, you've got the average speed, you have the time. I remember some equation (;)) to calculate distance from speed and time. Try to think for a while and work it out.
    Oh, and please use the edit button.
     
  10. Jan 11, 2006 #9
    V = at
    s = (at^2)/2 for uniformly accelerated motion where
    v = elocity at time t, a = constant acceleration and s = displacement in time t
     
  11. Jan 11, 2006 #10
    so if the rock is dropped Vi=0 vf=9.8*2.4=23.52m/s?? what about 2b
     
  12. Jan 11, 2006 #11
    You got that right. As i said, you can use the average speed for 2b. Then you have got a time and a velocity. Now, work that out on your own. If you are doing acceleration now, I am sure you have worked with constant velocity before.
     
  13. Jan 11, 2006 #12
    no i havent before this chapter was vectors and first chapter was conversions just give me the equation
     
    Last edited: Jan 11, 2006
  14. Jan 11, 2006 #13
    In that case your book is crap, but i guess that is not the issue here. Still, I will not give you the equation. I will give you hints.

    velocity is defined as m/s, or distance/time. Now get some math working on that one (5=10/2 might help you).
     
  15. Jan 11, 2006 #14
    I need to print out a table with all the conversion factors
     
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