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Relative velocity homework

  1. Nov 23, 2004 #1
    An object is thrown directly up (positive direction) with a velocity (vo) of 20 m/s and do= 0. How high does it rise (v = 0 cm/s at top of rise). Remember, acceleration is -9.8 m/s2.

    d =_______________m


    this is the problem I was given

    I think im to use d= do + vot + 1/2atsquared

    do=o
    and vo=20m/s
    but no time is given
    and v=0

    i realy don't know what to do
     
  2. jcsd
  3. Nov 23, 2004 #2

    arildno

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    You're quite right; that is the equation you need to determine the distance with, ONCE YOU KNOW THE TIME!!

    You are also right in saying the velocity is 0 at the top; hence, you CAN DETERMINE THE TIME TO GET THERE by using an equation which tells you what the velocity is at a given time:
    In general, we have:
    v(t)=v0-at (RIGHT?)
    So at T, at the top, v(T)=0, that is:
    0=v0-aT
    From this equation you can solve for the time T.
    Use that time value to determine how high you have risen (using the distance formula)
     
  4. Nov 23, 2004 #3
    A way to solve this particular problem is to consider that all the kinetic energy at time=0 will be converted into graviatational potential energy when the object reaches its highest point.

    kinetic energy is 1/2 m v^2

    potential energy is mgh
     
  5. Nov 23, 2004 #4
    You could also use [tex]v_f^2=v_0^2+2ad[/tex], since you know that at the apex of the projectile's motion the velocity is 0 m/s. That saves you a little bit of time since you don't have to calculate any extraneous variables.

    [Edit: just realized my way is pretty much exactly the same as ceptimus's.]
     
    Last edited: Nov 23, 2004
  6. Nov 24, 2004 #5
    ok

    don't I have to figure d not v(t)
    i still don't understand
     
  7. Nov 24, 2004 #6
    Yes, you need to figure out d. Since you know that the particle's velocity is 0 when its height = d, you can use the equation that nolachrymose posted, setting the left hand side = 0 and then solving for d since you have the rest of the quantities.
     
  8. Nov 29, 2004 #7
    is this it

    is it 419.6
     
  9. Nov 30, 2004 #8
    hello

    is this it I've not got a reply
     
  10. Nov 30, 2004 #9

    HallsofIvy

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    And you aren't going to get many replies. Show us what you have done!
    It is very difficult for us to give hints if we don't know what you know and what formulas you have to work with. I thought of several different ways to do this problem, some involving calculus and differential equations!

    (No, "419.6" isn't the answer. For one thing, the answer is a distance and must have the correct units. Even if you mean "the object will rise to a height of 419.6 meters", that is not even close. How did you get that?)

    You were told by arildno that v(t)= v<sub>0</sub>- 9.8t which is v(t)= 20- 9.8t since the initial speed is 20. Don't say "I don't want to find v(t)" that's not the point. In order to use d= -4.9t<sup>2</sup>+ 20t, you need to know the height. arildno's point was that an object will go up as long as its speed is positive, be coming down when its speed is negative. AT THE HIGHEST POINT, its speed will be 0. You can use
    v(t)= 20- 9.8t= 0 to find t AT THE HIGHEST POINT. Once you know that, put that value of t into d= -4.9t<sup>2</sup>+ 20t to find d.
     
  11. Nov 30, 2004 #10

    arildno

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    tman:
    My response was fashioned in the way it was in order to utilize your own, suggested approach (i.e, use of the distance formula).

    However, as HallsofIvy pointed out, you cannot SOLELY use that equation, because you need to find the actual time spent to reach the maximal height.
    That's where the velocity equation comes in; armed with the knowledge that the velocity MUST be zero on the top, the velocity equation may be used to detemine the time.

    I would, however, emphasize that it would be beneficial to your understanding to compute the answer in two ways:
    1) How I have proposed to solve the problem (on basis of what you suggested).
    2) Ceptimus' energy conservation argument.
     
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