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Projectile motion problems

  1. Oct 15, 2007 #1
    1. The problem statement, all variables and given/known data
    two problems. i am confused about their processes. they are almost the same but some things look strange.

    problem 1:
    A cannon on a level plain is aimed 50deg above the horizontal and a shell is fired with a muzzle velocity of 360m/s toward a vertical cliff 950 m away. how far above the bottom does the shell strike the side wall of the cliff?

    problem 2:
    A world series batter hits a home run ball with a velocity of 40m/s at an angle of 26deg above the horizontal. A fielder who has a reach of 2.2m above the ground is backed up against the bleacher wall which is 120 m from home plate. The ball was 1 m above the ground when it was hit. How high above the fielders glove does the ball pass?



    2. Relevant equations
    the kinematic equations.


    3. The attempt at a solution
    problem 1:
    Vox=360cos50deg=231.403m/s
    Voy=360sin50deg=275.78m/s
    time of flight to the cliff
    x=Voxt
    950m=231.403t
    t=4.1054s

    final vertical velocity at impact with cliff
    Vfy=Voy+ay*t
    =275.78+(-9.81)*(4.1054)
    =235.51m/s
    QUESTION: This Vfy is a positive number. Does this mean that the shell is still rising at this point? Had there not been a cliff it would have gone higher passed this point right?

    Height of the point of impact:
    Vf^2=Vo^2 +2aS
    Here i put Vf^2=0 because i want the displacement from the point of impact to the ground where Vf will be 0. and Vo will be the Vfy of the last step.
    a=-9.81m/s^2
    and the answer i get is 2.827km


    PROBLEM 2:
    Vox=40cos26=35.952m/s
    Voy=40sin26=17.5348m/s
    Time of impact with bleacher wall
    x=Voxt
    t=3.3378secs

    final vertical velocity at impact with bleacher
    Vfy=Voy+ay*t
    =17.5348+(-9.81)(3.3378)
    =-15.21m/s


    QUESTION: I get a negative value here. Does this mean that the ball is descending?

    Height of the point of impact:
    Vf^2=Vo^2 +2aS
    Here i put Vf^2=0 because i want the displacement from the point of impact to the ground where Vf will be 0. and Vo will be the Vfy of the last step.
    a=-9.81m/s^2
    and the answer i get is 11.7912m.

    11.7912 - (2.2-1) = 10.5912m

    Can someone see if I have done these questions correctly? thank you for your time.
     
  2. jcsd
  3. Oct 15, 2007 #2
    anyone?
     
  4. Oct 15, 2007 #3
    I got a different answer for the first one. After you got the time and the y velocity, is there any reason you didn't use this equation:

    [tex]y_2 = y_1 + v_1t + \frac{1}{2}at^2[/tex]

    It seems like this is all you need, instead of finding the final velocity then the distance. But if I've missed something, please let me know.

    [tex]\soutI haven't worked on the second yet.[/tex]

    O.k., for the second one I got a different answer than yours too. I re-did the first one using your formulas, and I got the same answer as I had previously calculated. I think there's a flaw in your thinking. For [tex]v_f[/tex] you cannot just change it to 0 because of your reasoning. For example, in the baseball problem, the ball is still moving after it passes over the fielder. The only time the velocity is zero in that case is when the ball is at the top of the curve. You can get the correct answers using the different formulas you used if you use the correct values for each variable.
     
    Last edited: Oct 15, 2007
  5. Oct 15, 2007 #4
    can you walk me through this equation?
    What is y2 and what is y1?
    Vi is initial velocity? is this Vix or Viy? or just the Vi?

    I haven't seen an equation setup like this.

    If I did it by getting the final vertical velocity at the point of impact and then the distance? is this wrong for some reason? You got a different answer so there must be something that I don't know. Tell me why that wouldn't work?

    Thanks for responded.
     
  6. Oct 15, 2007 #5
    Chocokat seems to be correct. Once you have the time and the initial y velocity you can just use the displacement formula.

    I'm not really sure why you tried to find the final velocity and use that to find the height. I guess it should technically work but I think the problem is that you only found velocity in the y direction at that point and didn't include the x direction. Which would make a vector with a magnitude equal to [tex]\sqrt{V_{ox}^{2}+V_{oy}^{2}}[/tex]

    Edit: Just saw your reply after I posted.

    The formula that Choco posted is a displacement formula. It is normally solved as D=Vot+1/2at^{2} then the starting height can be added when necessary.

    y2 is just the final y position where y1 is the initial y position.

    Vi is initial velocity. It is going to be either Vix or Viy depending on what you are trying to find, it won't ever be Vi (unless the Vi is vertical or horizontal).

    Basically you will use his formula or D=Vot+1/2at^{2} and plug in what you know to find the height.

    Since the x acceleration is usually 0 in beginning problems, you will see that the formula breaks down to x=voxt, the formula that you have already used once.
     
    Last edited: Oct 15, 2007
  7. Oct 15, 2007 #6
    [tex]y_2 [/tex] is the final y
    [tex]y_1[/tex] is the original y, in problem #1 it is 0

    Your formulas work, as long as you use the actual final velocity (235.551) in the
    [tex]vf^2 = vi^2 + 2aS[/tex]

    it's just more work (and therefore more room for error) than the one I used. But, if you don't have that one (as well as the corresponding x and general one) you may not want to use it.

    Just re-do your math, the other calculations look good.

    EDIT:
    should be 'her' although they really aren't 'my' formulas, I share them with everyone...
     
    Last edited: Oct 15, 2007
  8. Oct 15, 2007 #7
    ok now for number 1 i am getting 1.0495km
    number 2 i am getting 2.6815 m

    are these teh values you are getting? thanks.
     
  9. Oct 15, 2007 #8
    I got the same answer for #1. For #2 I got 2.735m, but the difference may be due to little rounding differences (or maybe I entered something wrong) between your and my calculations. Double check your calculations, and if you come up the same, then you're probably fine.
     
  10. Oct 15, 2007 #9
    Yes those are the values I get.

    PS: Sorry if I offended you, I shouldn't assume gender, this being the internet and all.
     
  11. Oct 15, 2007 #10
    I wasn't offended, just giving you a hard time.

    Pooface - Since Matty got the same answer as you, I think you're good to go! Good luck.
     
  12. Oct 15, 2007 #11
    thank you very much for your help! both of you! I really appreciate it.
     
  13. Aug 27, 2011 #12
    *When I write in capital letters it means I want to emphasize a point. I am not yelling at you. : )

    I think that the 1049 meters is actually the height of the projectile AT IMPACT; This is NOT the maximum height of the projectile. What you did was include the velocity of impact in your calculations. This is what you did:
    [(Velocity of Impact)^2 - (Initial Velocity)^2 ] / 2(-9.81 meters per second squared)
    [(235.51^2 - 275.78^2 ] / -19.62 mpss
    The maximum height of the projectile is also NOT 2 kilometers, it is 3876.38 meters or 3.876 Kilometers.
    Here are your calculations: *The part I ***___*** is wrong:
    0 = (***235.51^2***) + 2(-9.81)y
    Once again, you included the velocity of impact in your calculations where you should have included your initial velocity, which is 275.78 m/s. Your final velocity is zero because this is when the projectile stops moving.
    0 = (275.78^2) + 2(-9.81)y
    Then you would solve for y: y=3876.86 meters, or 3.876 kilometers
    Therefore, the projectile would have gone higher if there were no cliff
     
  14. Aug 27, 2011 #13
    You did a great job on problem 2, your work is correct:
    The time of flight is around 3.34 seconds. The velocity at impact is around -15.21 seconds. The reason for this negative sign is because at this time the ball is still moving upwards contrary to the force of gravity.
    Suggestion: It would help to contstruct a frame of reference.
    Physics Forum Kinematics.png
    *** See my attachment.

    Your calculation that the height at impact is 10.592 meters is also correct. I calculate the maximum height of the ball, which is 15.6712 meters.
    (Final velocity)^2 = (Initial velocity)^2 + 2 (acceleration due to gravity)(y)
    Final velocity is zero when object hits ground or catchers mit.
    0 = 17.5348^2 + 2(-9.81)y
    Solving for y gives me y = 15.6712 meters.
    Because the height at impact is around 10.6 meters, which is LESS than the maximum height, this means that the ball is still ascending at this point.
     
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