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Homework Help: Projectile Problems

  1. Oct 12, 2007 #1
    1. The problem statement, all variables and given/known data

    projectile intial velocity vo of 49 meters per second. angle theta intial is 76 degrees. projectile takes off and lands on the same horizontal plane. neglecting air resistance and using g=9.8 m/s squared. find

    a. range of projectile: 115.020m
    b. total time of flight: 9.70295
    c. time in which the projectile is farthest from its launch point: 4.85147
    d. farthest distance the projectile is from its launch point: 115.330
    e. the critical angle theta c such that for theta c is greater than or equal to theta which is greater than 0 the projectile is always moving farther away from its launch point and for theta greater than theta c there are times during the projectiles trajectory when the projectile is moving towards its launch point: not sure where to begin

    give six significant digits in all answers

    2. Relevant equations

    Vox = Vo Cos(theta); Voy = Vo sin(theta); 1/2t=Voy/ay; t=2(1/2t)
    dy=Voyt+.5ayt^2; dx=Voxt

    3. The attempt at a solution

    Vox= 49 cos76= 11.85417
    Voy= 49 sin76= 47.54449
    1/2t= 47.54449/9.8= 4.851479
    dy= (47.54449)(4.851479)+(.5)(-9.8)(4.851479)(4.851479)=230.6611-115.3306= 115.330
    t= 2(4.851479)= 9.70295
    dx= (11.85417)(9.702958)= 115.020

    I have been out of school for quite sometime now and didn't do so well on physics then. I believe I am close on the range and total time. I feel I am missing something for farthest point. I was assuming this meant the peak of its arc from the ground. But after reading it again I am not so sure. And Theta C I dont get since its an arc not an angle unless its talking about the angle of launch. Any hint in the right direction would be appreciated. Thanks.
  2. jcsd
  3. Oct 12, 2007 #2


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    EDIT: Sorry! thanks chocokat! silly of me... because dy is just 0 at the range anyway.

    Looks like all your answers are correct Brentavo7!
    Last edited: Oct 12, 2007
  4. Oct 12, 2007 #3
    Learningphysics - since the projectile goes up and comes back down, at 9.7### it'll be back on the ground (or at the same vertical place where it started), whereas, won't using the 4.85### it'll calculate the top of the arc, i.e. the farthest distance from the launch point?
  5. Oct 12, 2007 #4


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    Yes, you're absolutely right. :redface: Thanks!
  6. Oct 12, 2007 #5
    So at a projectile's max height, it is furthest from its launch point? (I am not in doubt that you are correct, I am just having trouble reasoning why).

  7. Oct 12, 2007 #6


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    I am in doubt that's correct, because I don't think it is. And I have no clue what ### means.
  8. Oct 12, 2007 #7
    No, the furthest point may or may not be the height, but in this case it is (based on my calculations, which I hope are correct), but just by a little.

    Height = 115.33m
    Horizontal distance = 115.02m
    Last edited: Oct 12, 2007
  9. Oct 12, 2007 #8
    Dick - sorry, I used ### as I only went out 3 places, but the original question needed 6, and I didn't bother to get the rest of the places.
  10. Oct 12, 2007 #9


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    Yes, you're right. I think I'll avoid posting on this thread for a little while... think I'm a little too tired tonight. :tongue:
  11. Oct 12, 2007 #10


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    That's ok. But at the farthest distance from the launch point the velocity should be perpendicular to a line joining it to the launch point. Can't be true at the top point of the trajectory if there is any horizontal velocity. Significant places are another matter and you are right that the difference is smallish. But the OP was asked for six sig figs.
  12. Oct 13, 2007 #11


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    So parts a and b are good. for part c)

    distance from the launch point d^2 = x^2 + y^2

    take the derivative with respect to time

    2d*dd/dt = 2x*dx/dt + 2y*dy/dt

    since dd/dt = 0 at the max...

    0 = x*dx/dt + y*dy/dt (giving that (x,y) is perpendicular to (dx/dt, dy/dt) as Dick mentioned)

    plug these in and solve for t... I think it will slightly more than 4.851479...

    then for part d), get x and y at the time you found in part c).

    For part e)... if the object is always moving away from the launch point... that means dd/dt always > 0

    ie x*dx/dt + y*dy/dt always >0... following Dick's suggestion from another thread, take equation x*dx/dt + y*dy/dt = 0... this reduces to a quadratic... find the range of the angle theta such that this has no real roots.
    Last edited: Oct 13, 2007
  13. Oct 13, 2007 #12
    Thank you all for the help. I now see that the height was not the farthest from the launch point. Using the formulas mentioned. I am now getting the farthest distance of130.055 meters at 5.38863 seconds. Critical Angle of 71.4856 degrees. Still not positive Im doing the angle correctly. Ive got to get some sleep though. Ill try again in the morning. Thanks again.
  14. Oct 14, 2007 #13


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    I'm not getting exactly the same answers (but close though). Can you show how you got them?
  15. Oct 14, 2007 #14
    I actually ended up cheating a bit. I used autocad to map out the curve and placed a circle from the launch point to the max height. saw where the curve intersected the circle again. Took the mid point of the two intersections as the farthest point from the launch point. This gave me a distance of 130.076m at 5.33971s. I would like to verify by formula. I am confused on dd/dt=0.
  16. Oct 14, 2007 #15


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    Oh I see... I think that's pretty cool that you were able to use autocad for that.

    The algebra is a bit cumbersome for the problem...

    starting here:
    d^2 = x^2 + y^2

    d is the distance of the projectile from the start point.

    you want to find the spot where dd/dt = 0. Generally the max or min of a variable occurs when its derivative is 0.

    so the first step is taking the derivative with respect to t, of

    d^2 = x^2 + y^2
  17. Oct 14, 2007 #16
    ok Im begining to grasp it. I seem to be reverse engineering though. if i use the formula
    d(dd/dt) = x(dx/dt) + y(dy/dt) and plug in the numbers I found. I get
    130.0765 * 130.0765 /5.33971 = (63.29792 * 63.29792 /5.33971)+
    (113.6366 + 113.6366 /5.33971)

    16919.9 / 5.33971 = (4006.627 / 5.33971) + (12913.28 / 5.33971)
    3168.693 = 750.3454 + 2418.349
    3168.693 = 3168.694

    As far as angle Theta c is concerned am I closer with the original 71 something degrees or is it more common sense of 67.5
  18. Oct 14, 2007 #17
    Thanks for the compliment. Ive been a fire protection engineer for 6 years now. We use Autocad extensively. Ive got Angle Theta C to be somewhere between 71.85 and 71.9 degrees. Does anyone happen to know the number out to six significant figures.
  19. Oct 14, 2007 #18


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    I recommend going through the algebra here... it's actually not too bad now that I go through it... although your numbers are close, they aren't close enough...

    x*dx/dt + ydy/dt = 0

    [tex](vcos(\theta)t)(vcos\theta) + (vsin(\theta)t - (1/2)gt^2)(vsin(\theta)-gt) = 0[/tex]

    I recommend going through it without plugging in numbers till the very end.

    first thing you can do is divide by t...

    you'll get a quadratic in terms of t...

    use the quadratic formula to solve for t...

    so this actually takes care of both the time to get the maximum distance... and also the angle... when the discriminant< 0 there is no solution... and that gives the critical angle.

    I'll verify your expression when you get it.
    Last edited: Oct 14, 2007
  20. Oct 16, 2007 #19
    R = Vo2 Sin(2θ)
    g ​
    R = (49m/s)2 Sin(2(76o))
    R = 115.02053288253m​

    TFinal = 2 Vo Sin(θ)

    TFinal = 2 (49m/s)(Sin76o)
    TFinal = 9.702957263s​

    Y = -(1/2)(g)(t2) +Vo Sin(θo)t
    = -4.9 t2 + 47.544490587524t – 115.33054006261

    then t => -b +/- √((b2) – (4)(a)(c))
    t = 4.85147863138s​
    H = Vo2 Sin2 (θ)

    = (49m/s) 2 Sin2 (76)

    H = 115.33054006261m​

    X = Vo Cos(θ)t
    X = 57.510266441273m

    Y = -(1/2)(g)(t2) +Vo Sin(θo)t
    Y= 115.33054006261m

    θc =ArcTan(Y/X)

    θc = 63.496586096704o​

    Anyone want to check that? I think that’s right.
  21. Oct 16, 2007 #20
    some of the ^2 came out as just 2 but the answeres are still correct.
  22. Oct 16, 2007 #21
    Hey Andy check c and d. The height isnt the farthest point. use the equation in post 18.

    My math may be wrong but Im getting:

    (Vox^2 + Voy^2) - (14.7t*Voy) + (48.02t^2) = 0
    2400.9998 - 698.904t + 48.02t^2 = 0
    Im getting t = 5.55736. That right?
  23. Oct 16, 2007 #22
    hmm at 5.55736s wouldn't that place the object at (x,y) (65.87790622,112.8890244) versus (57.510266441273, 115.33054006261m) hmm I will go back over it later tonight. to tired to try and do any more math right now lol.
  24. Oct 16, 2007 #23
    The math seems to work for 5.55736 but I also get 8.99707 out of the quadratic.
  25. Oct 16, 2007 #24
    use my numbers and see if you get the same answere.
  26. Oct 16, 2007 #25
    with t= 4.85147 the distance from the launch point is 128.87m using x^2 + y^2 = d^2
    at t= 5.55736 the distance from the launch point is 130.705m
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