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Physics answer check

jai6638jai6638 Registered User Posts: 1,025 Member
edited January 2008 in Engineering Majors
Hey there..

I was given a question in a quiz for my Physics 1 class and I solved it. However, it was marked wrong. Could anyone please highlight why it is wrong?

Q) At a hockey game, you throw your hat into the air when our team scores a point. If you throw it upwards at an angle of 35 degrees with respect to the horizontal and speed of 11 m/s:

D) Assuming you are on level ground, how far away from you does it land ?

I found the time it takes to get to the highest point to be ( .6438 s and its maximum height to be .32 m ) in the earlier questions that I have not typed in this post.

To find this answer, I used the Range equation, R=((V0)^(2) * sin(2theta))/(g)
R= (11)^(2) * sin ( 2*35) / ( 9.8 )
R = 11.6 m.

Apparently, the answer is 5.8m but I'm not sure why it should be half of my answer. The range gives you the distance it covers in the X dimension so why is it half?!

Thanks
Post edited by jai6638 on

Replies to: Physics answer check

  • xjisxjis Registered User Posts: 458 Member
    i'm apparently getting the same answer (11.6).

    you already figured it takes .6438s to reach its highest point,
    so it takes 1.2876s total to fall down.

    Vx=11*cos(35)
    t=1.2876

    X=Vx*t
    =11*cos(35)*1.2876
    =11.6

    hmm.
  • lil_killer129lil_killer129 Registered User Posts: 4,706 Senior Member
    I think I have a reason. The range equation assumes your initial position and final position are at the SAME elevation. When the dude throws the hat, the hat obviously does not start from level ground, but yet it hits the ground (distance you calculated). I hope this helps.
  • xjisxjis Registered User Posts: 458 Member
    lil_killer129/

    if you assume that, it would travel even farther, not less.

    jai6638/

    it's not related to the current problem, but how did you get a maximum height of 0.32m? it should be 2.03m

    t=0.6438s (time to get to the maximum height)
    V_y0=11*sin(35)=6.31m/s

    Y=V_y0*t-gt^2/2
    =6.31*0.6438-(9.8)(.6438)^2/2
    =2.03
    =2.02m

    ...or maybe it's been too long since i've taken intro physics...
  • PSLaplacePSLaplace Registered User Posts: 63 Junior Member
    I also get 11.6 m.
  • lil_killer129lil_killer129 Registered User Posts: 4,706 Senior Member
    lil_killer129/

    if you assume that, it would travel even farther, not less.
    I didn't say it was less.
  • xjisxjis Registered User Posts: 458 Member
    lil_killer129/ jai6638 said "correct answer" seems to be the half of what he's got.
    Therefore, your assumption does not help because it would make it go farther instead.
  • ShacklefordShackleford Registered User Posts: 618 Member
    I also am unsure as to why 11.6 m is incorrect.
  • VorfieldVorfield Registered User Posts: 53 Junior Member
    The question didn't mention the plasticy board 5.8 meters away from you to prevent the hockey players from smashing into you or getting you involved in one of their brawls.

    Taking this fact into consideration, the range of the hat is 5.8m.
  • awesomoawesomo Registered User Posts: 139 Junior Member
    Dude, I would be ****ed if that was the real answer
  • lil_killer129lil_killer129 Registered User Posts: 4,706 Senior Member
    lil_killer129/ jai6638 said "correct answer" seems to be the half of what he's got.
    Therefore, your assumption does not help because it would make it go farther instead.
    Where are you getting this from? Where in my past posts have I said the correct answer was half of what the other dude got? I simply restated the facts about what the range equation means, which obviously is being interpreted incorrectly here. I thought reminding us of how the range equation is suppose to be used would provide some guidance into achieving the correct answer.
  • ShacklefordShackleford Registered User Posts: 618 Member
    Well, the initial height of the projectile is not given.

    Well, it does say, "assume you are on ground level."
  • PSLaplacePSLaplace Registered User Posts: 63 Junior Member
    lil_killer129 - What xjis is referring to is the "answer" of 5.8 m given by the topic creator in his first post. The topic creator is trying to figure out why it is 5.8 m and not 11.6 m.

    Your assumption would cause the object to travel further, not less, and is therefore not a suitable explanation. The explanation is, unless other info has been omitted, 11.6 m is correct, and TC should submit the problem for regrading.
  • ShacklefordShackleford Registered User Posts: 618 Member
    I concur. Maybe there was a typographical error somewhere.
This discussion has been closed.