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Projectile Motion/Vertex small problems.

  1. Sep 3, 2008 #1
    A Couple problems im having trouble with.

    1.A ball is thrown with a speed of 30 m/s and at an angle of 40 degrees. How high does it go above the cliff to the nearest tenth of a meter?

    Ok, well i worked that portion out...and im fairly sure it's 19 meters exactly....but feel free to double check, but this portion troubles me.

    In the previous problem, if the ball hits the ground 4.2 seconds later, how high to the nearest tenth of a meter is the cliff?

    2.A baseball is thrown horizontally off a cliff with a speed of 14 m/s. What is the horizontal distance, to the nearest tenth of a meter, from the face of the cliff after 2.8 seconds?

    That parts easy..39.2 meters...but once again theres a hard part i can't get..

    In the previous problem, to the nearest tenth of a meter, how far has it fallen in that time?


    3.A vector has x component 8.2 and y component 7.7 . What is its magnitude to the nearest tenth of a unit?

    That ones suppose to be an easy one...but i don't know how to find magnitude....i feel dumb.

    What is the direction of the vector in Problem 3 to the nearest tenth of a degree? (Report your results in the format -180 to 180.)

    I don't understand even how to set this up.

    sooo please please help me out a bit, just if someone could show me how to set up and work through these that'd be fantastic. I didn't show attempts at the problems, b/c the ones i knew how to attempt i got all the way through..the others i have no idea about..im really bad with vectors.
  2. jcsd
  3. Sep 3, 2008 #2


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    Welcome to PF.

    I wonder do you have a picture? Or could you describe what the problem is a bit more explicitly? Standing on the cliff throwing off? Or toward a cliff? As you can imagine that would make a difference.
  4. Sep 3, 2008 #3

    sorry....and with that picture i've given you all the information i have.
  5. Sep 3, 2008 #4


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    OK. I get the first part within a few hundredths.

    The second part is solved by first determining how long it took to reach that height. And then subtract that time from total time and you know how long it took to hit the ground from its peak.

    Subtract the height above the cliff and you have the height of the cliff.
  6. Sep 3, 2008 #5
    ok, when i divided 19.3(30sin40) by 9.8, i got Time as 2 seconds...is that the time it takes to reach that height?
  7. Sep 3, 2008 #6


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    Welcome to PF!

    Hi CDink! Welcome to PF! :smile:

    Just draw the triangle …

    then use Pythagoras for the magnitude, and tan for the angle. :smile:
  8. Sep 3, 2008 #7
    Re: Welcome to PF!

    Ok...im sure SOCATOA or whatever would come in handy but i never learned it....so is it 8.2tan7.7 or 7.7tan8.2?

    also i thought thats what it was...but i didn't seem to do it right or something...i added 7.7 and 8.2...did i do it wrong?
  9. Sep 3, 2008 #8


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    I think we have uncovered the source of your confusion.

    The link below offers some formulas that describe how to find the value of distance, time, velocity or acceleration when you are given some but not all of the variables.

    As to your attempt you are no longer interested in the initial velocity since you already have the height. In a 1 dimensional problem (remember we are only considering the up and down components since that's all that's asked for) you can determine time from distance with:
    x = 1/2 a t2

    You know x, you know a, so ...

    Once you have that time you can determine how much time remains. And then this time, knowing time and acceleration solve for the distance back down to the ground below.
  10. Sep 3, 2008 #9
    Ok, well i still got 2 seconds......and i know

    Starting v=0(b/c it stops at the top right?)
    V=19.6? i think i worked it right

    but i need to find the total x, correct? soooo


    did i set that up correctly? well apparently not since i got the answer to be 19.6 meters...making the cliff .6 meters....hmmm what'd i do wrong.
  11. Sep 3, 2008 #10


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    Hi CDink! :smile:

    I think SOCATOA is more confusing than learning it straight …

    for tan, just remember that tan (tangent) is the same as slope!

    So tan = opp/adj.

    Also, whatever follows "tan" must be an angle … like tan7º, or tan(π/4).

    7.7 and 8.2 are lengths, not angles, so they're not going to follow "tan", are they? :wink:
  12. Sep 3, 2008 #11
    Re: tan

    well than how do i use it to find the second portion of that problem? or do i even?...i found the magnitude however...seems to be 11.3
  13. Sep 3, 2008 #12


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    OK, I got 1.96 seconds. And there are 4.2 total seconds that means even using your number of 2 that there was 2.2 seconds for it to drop. That yields more than .6 meters additional height from the cliff.

    Btw, doing it your way on the downward trip your final velocity will be a * t = 2.2 (9.8) = 21.56 not 19.6.
    Last edited: Sep 3, 2008
  14. Sep 3, 2008 #13
    ok, well now i don't know where to plug in that 2.2....but i now know what it's from it reaches the top at 2 seconds, and the bottom of the cliff at 4.2....but my V2 equation doesn't even bother with time..
  15. Sep 3, 2008 #14


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    What is your aversion to using x = 1/2 a t2

    Btw, doing it your way on the downward trip your final velocity will be a * t = 2.2 (9.8) = 21.56 not 19.6.
  16. Sep 3, 2008 #15


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    Actually your way does use t. You need t to figure final velocity and that is given by a * t.
  17. Sep 3, 2008 #16
    wow....sorry, ugh i over think this stuff.......ok, im getting to 23.7 meters as the Total height making the height of the cliff 4.7 meters? awfully small cliff..im sorry im so stupid with this.
  18. Sep 3, 2008 #17


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    A meter is 3 feet that's nearly 15 feet.
  19. Sep 3, 2008 #18


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    Having just done this in the first question, then this should be pretty simple now right?
  20. Sep 3, 2008 #19
    so my answer is correct? if thats the case than 1 down 2 to go.
  21. Sep 3, 2008 #20

    umm yeah soo i just..x=14*7.8?
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