1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this okay?

  1. Oct 28, 2004 #1
    I was wondering if I could post some problems from my physics homework that I did so that it can be checked by you guys. So is it okay?
     
  2. jcsd
  3. Oct 28, 2004 #2
    sure, go ahead
     
  4. Oct 28, 2004 #3
    here's one I'm not sure about. More specifically, part B.

    2.) A canoe is paddled at 4 km/h directly across a river that flows at 3 km/h, as shown in the figure. (a) What is the resultant speed of the canoe? (b) How fast and in what direction can the canoe be paddled to reach a destination directly across the river?

    (a) resultant^2 = (magnitude of horizontal vector)^2 + (magnitiude of verical vector)^2

    resultant^2= (3 km/h)^2 + (4 km/h)^2

    resultant= 9 km/h + 16 km/h

    25=25

    (25)^1/2= 5 the resultant is 5 km/h

    (b) The canoe needs to be paddled perpendicular to the vertical component and will neccessarily be doing so at 5 km/h as shown in (a).
     
  5. Oct 28, 2004 #4
    Yeah, a is correct

    For b, you need to have a velocity that goes exactly against the current of 3m/s

    So, you break it up into...
    4^2=3^2+v^2
    v is the vertical velocity you are traveling at to get across the stream
     
  6. Oct 28, 2004 #5
    thanks. for this one, I know I'm wrong:

    5.) Students in a lab measure the speed of a steel ball launched horizontally from a table top to be 4.0 m/s. If the table top is 1.5m above the floor, where should they place a 20 cm tin coffee can to catch the ball when it lands?

    given: s= 4.0m/s, d= 1.5

    formulas: resultant^2 = (magnitude of horizontal vector)^2 + (magnitiude of verical vector)^2

    (4.0m/s) ^2= x^2 + 1.5^2

    16m^2/s^2=x^2 + 2.3m^2

    16m^2/s^2-2.3m^2=x^2

    13.7 m^2= x^2

    3m=x



    thanks.
     
  7. Oct 28, 2004 #6
    Find time it took for the ball to drop:

    d=.5at^2
    1.5m=.5(9.8m/s^2)t^2
    t= ....?

    Then find how far it would travel after that many seconds
    d=vt
    v=4m/s(t)
     
  8. Oct 28, 2004 #7
    since it is launched from the table, it will be a projectile

    you know the horizontal velocity v1, what is tha cceleration HORIZONTALLY, thus you can figure out the final velocity and thereby find the horizontal ditance it travels horizontally.

    When they say the tin can is 20cm what do they mean 20cm? Do you know or are you suppposed to guess??

    In any case the tin can should be placed +/- 10cm (if the diameter is 20cm) fro mthe point where tha ball will fall so the ball can fall to the middle of the can.
     
  9. Oct 28, 2004 #8
    if t=.5 s, then d=2m.

    I don't understand how the vertical drop velocity has anything to do with the magnitude of the horizontal component.
     
  10. Oct 28, 2004 #9
    Where in the can the ball lands is irrelavant to the question. I think they just added the 20cm part for the sake of challenging the student.
     
  11. Oct 28, 2004 #10
    The vertical velocity has NOTHING to do with the x velocity. However, they are in the air for the same amount of time. So you find the time it took for the ball to drop, you have the time for the ball to travel in the x direction. Hence, d=vt
     
  12. Oct 28, 2004 #11
    Ah!!! This reminds me of an example in our text book:
    If you drop a bullet and at the same time shoot a bullet (with a gun), which one of the two bullets will land on the ground first, given that the ground is exactly parallel to the initial starting point of the motion of the bullets?
    Both will reach the ground at the same time.

    I really don't understand why. Isn't one ball traveling on a curved path while the other on a path vertical to the ground? Wouldn't that make the ball being shot then take a longer time to touch the ground?
     
  13. Oct 28, 2004 #12
    Gravity attracts everything to the ground at the same acceleration, moving or stationary.

    The other ball looks like it is travelling a curved path because it is travelling horizontally and vertically at the same time.
     
  14. Oct 28, 2004 #13
    what does it mean for something to go on a curved path then? Isn't it a combination of horizontal and vertical components? I really don't get this.

    (I understand g is constant for earth. )
     
  15. Oct 28, 2004 #14
    Check this out.
     
  16. Dec 20, 2004 #15
    gravity has no effect on horizontal velocity.
    Hope this helps clear the confusion.
     
  17. Dec 20, 2004 #16
    The result shows that the vertical and horizontal components of motion do not directly affect each other. You can always separate the projectile's motion into two orthogonal sets of equations. Note that while you may need to get the time something happens in the vertical motion to see what happens simultaneously in the horizontal motion, the equations are still independent.
     
  18. Dec 20, 2004 #17

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I would interpret the "20 cm. can" as meaning the can is 20 cm. TALL. You need to position the can so the marble is 20 cm.= 0.2 m. above the floor when it gets to the can.

    That just means that you can use 1.5- 0.2= 1.3 m as the height instead of 1.5m.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Is this okay?
  1. Is this proof okay? (Replies: 3)

Loading...