1. Limited time only! Sign up for a free 30min personal 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!

Trajectory Help

  1. Sep 16, 2009 #1
    I found a similar question here from a couple years ago, but didn't see the whole problem or a complete explination of the solution so I wanted to repost and hopefully get some insight to help with figuring this whole thing out.

    Problem:
    A projectile is shot from the edge of a cliff 125m above ground level with an initial speed of 65.0 m/s at an angle of 37degrees above the horizontal.

    a) Determine the time taken by the projectile to hit point P(landing point on ground).
    b) Determine the range of the projectile as measured fro the base of the cliff to point P. At the instant just before the projectile hits point P, find
    c) the horizontal and the vertical components of its velocity,
    d) the magniture of the velocity, and
    e) the angle made by the velocity vector with the horizontal.
    f) Find the maximum height above the cliff top reached by the projectile.

    I have looked at kenematic equations and think that could help part of this, but. Just not sure where to start and how to get there...
     
  2. jcsd
  3. Sep 16, 2009 #2
    I guess I should have added that this is for me to help my daughter out with a problem she's struggling with. I absolutely LOVED physics when I was in high school; and she's got that same bug. But, I dont' remember ever doing anything this involved with physics when I was in school. Or, maybe I just blocked this part out?? :D

    So, while not exactly looking for an answer; I am just looking for hot to explain to her how to get there.
     
  4. Sep 16, 2009 #3

    rl.bhat

    User Avatar
    Homework Helper

    Try to find the equation of the projectile motion.
    In the problem, y,initial velocity and angle of projection is given.
    Projectile motion is the combination of horizontal and vertical motion.
    Find the components of the velocity in the horizontal and vertical direction.
    From the vertical component you can find the time of flight.
    From this time of flight you can find the range of the projectile.
     
  5. Sep 16, 2009 #4
    Okay. but from what I have been able to gather reading up on this problem:

    A) it is not stated in the problem, but should I ignore air resistance? (Considering this is a Junior HS level Physics class (although an AP class) I would assume the answer here is yes.)

    B) Ignoring air resistance, the V-horizontal remains constant ... correct?
     
  6. Sep 16, 2009 #5

    rl.bhat

    User Avatar
    Homework Helper

    No. Horizontal component remains constant.
    Acceleration due to gravity acts on the vertical component.
     
  7. Sep 16, 2009 #6
    Good Catch.. :) I mis-typed that, but yes, horizontal component remains constant, vertical changes due to gravity. :D

    I'm trying to figure the Vx and Vy now.. Sin/Cos/Tan...arrrggh I never did like them words! LoL
     
  8. Sep 16, 2009 #7

    rl.bhat

    User Avatar
    Homework Helper

    Vertical component is vsinθ and horizontal component is vcosθ.
     
  9. Sep 16, 2009 #8
    I think I'm on the right track.. Thx for the help!
    Just to verify, here's what I got so far.

    Vox=51.9m/s
    Voy=39.1m/s
    t (peak of trajectory)=3.98s
    dy (y-displacement)=78m

    From this I know that it travels 7.96s until it is at the same hight as it left. I need to figure the rest since it started out 125m high. Therefor it falls 203m total.

    Hmmm.. I think I am on to figuring out that part.. should be just as easy; as well as total distance traveled. I believe what will trip me up is results for C, D and E.

    Any hints there? :D
     
  10. Sep 16, 2009 #9
    not sure of my equations here, so...

    If I am right and the max hight is 203m (inital 125m + y-displacement of 78m)then:

    203=(.5)(a)(t^2)
    203=(.5)(-9.8)(t^2)
    207.9=(t^2)
    14.4s = t (approximate)

    So, overall flight time = 18.38sec

    I'm fairly sure of the time to peak of trajectory (3.98), but if the fall time is wrong everything else I do after this point will be wrong... Can someone confirm or correct me here?
     
  11. Sep 16, 2009 #10
    I'm not an expert in the subject, but I can't get the same answer for the overall flight time. The flight time only depends on the y-axis, so if you use V0y(the same you calculated) and d = V0y t - 0,5 g t^2 and use the fact that d = -125 then you get a different answer. Is this approach for calculating time wrong?
     
  12. Sep 16, 2009 #11
    Wouldn't (d) be positive? That's how I got the fall time of 14.4.

    But, if I take this as a -125, then add the rise of 78 and (1/2a), gives 51.8
    then the fall time is 7.2s
    This is a lot closer to what I would think it should be..

    HOWEVER: if my time to peak of 3.98s is right, that means it gained (+y) 78m in just under 4s
    so, a fall of 203 taking just over 14.4 is pretty close to proportional to the rise
    (78/4 approximatly = to 203/14 )(19.5 and 14.5)
    vs a figure totally unproportional
    (78/4 not even close to 203/7)(19.5 and 29)

    I think I should be able to deduce from this that the reason the 14.4 fall time is not 'identically' proportional is because it is falling further (1.5x further) therefor it has more time to build up -Vy.

    And here is why I am no longer in physics.. LoL.. I love to overthink things. :D
     
  13. Sep 16, 2009 #12
    -125 = Voy t - 0,5gt^2 gives t = 10,4. d is negative because if you set y = 0 at the level which you threw from, then the ball lands 125 meters below that. Think of it as a coordinate system.

    English is not my first language so I have difficulty understanding your explanation but 3,98 seconds to the top sounds right if Vy = Voy - gt, and you set Vy = 0.
     
  14. Sep 16, 2009 #13
    Something is different here...
    You have Y=(Voy)(t)-(0.5)(g)(t^2)

    Everything I ahve seen so far is:
    Y=(Voy)(t)+(0.5)(g)(t^2)

    What is the difference between these two? I have not seen this formula expressed with the - sign
     
  15. Sep 16, 2009 #14
    The reason is that the direction of Voy is straight up in the beginning, whereas gravity always works downwards (Everything falls down).

    Vy will constantly decrease until it reaches 0 (at the top of the trajectory), and will then have a negative velocity (it falls in the opposite direction)
     
  16. Sep 16, 2009 #15
    okay. are you using 9.8m/s/s for g or are you using -9.8m/s/s?
    Maybe that is the difference? The papers that I have been reading show to use a -9.8.


    Back to my original problem though; looking at result "C"
    the vertical and horizontal velocity at impact. The vertical is going to be derived from (t), but the horizontal should be the same at the end as it was at the start. (g) should not effect horizontal velocity... correct?
     
  17. Sep 16, 2009 #16
    You've got it right, gravity is -9,8 so putting it into gt^2 / 2 will -gt^2 / 2, thus d = Vot - gt^2 / 2. The standard formula is d = Vot + at^2 / 2
     
  18. Sep 16, 2009 #17
    Anden..
    Okay - I'm with you now.
    However, lets go back to your original solution;
    -125 = Voy t - 0,5gt^2 gives t = 10,4

    But did you miss - or am I missing the extra 78m it rose before falling in your equation? Should not the -125 be instead 203. If so, then my 14.4s is correct. I need how long it took to fall the entire 203m; so total flight would be total rise time + total fall time.
     
  19. Sep 16, 2009 #18
    Actually, that single equation covers the entire vertical trajectory. If you put t = 3,98 into d = 39,1t - gt^2 / 2 => d = 78 m.

    We know where the ball is going to end, 125 m below the original point, so we set d = -125. Then we get a second-degree equation, which when solved yields t = 10,4.

    If you try plotting the graph of the equation you will see how the curve first rises until t = 3,98 and d = 78 and then start to descend until t = 10,4 and d = -125 or until t = 14,4 and d = -203. When you use 203 the cliff becomes 203 meters high instead of 125.

    This might be a bit confusing, but see d as distance from the Original point.
     
  20. Sep 16, 2009 #19

    rl.bhat

    User Avatar
    Homework Helper

    In such problems you have to consider displacement not the total distance covered.
    Displacement is a vector whose magnitude is the distance between the final position and the initial position, and the direction is from initial position to the final position.
     
  21. Sep 16, 2009 #20
    I think I'm confused more now! :D

    I talked w/my daughter tonight, and the correct answer for total flight time is 10.4. Anden you are right.. But this is what confuses me. I will try to find the website I was reading on how to figure this, but from my knowledge, I would think that Voy would be 0.0m/s. Because at the top of the trajectory the vertical velocity is zero.

    But reading the explination above... I understand the distance traveled (y) on a graph would be to a -125. Makes total sense. But that 78m increase. where is it figured in?
    If the cliff was 203m high, the trajectory would still take it up 78m. Maybe that is why I am lost... I cannot see where the 78m increase has gone.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trajectory Help
  1. Trajectory HELP? (Replies: 1)

  2. Trajectory help (Replies: 3)

Loading...