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Homework Help: Projectile Motion Range (please Help)

  1. Oct 31, 2006 #1
    Question: How far will a stone travel over level ground if it is thrown upward at an angle of 31.0 degrees with respect to the horizontal and with a speed of 11.0m/s?

    What is the maximum range that could be achieved with the same initial speed?

    I have tried to solve this with no sucess what i have done is Drawn a triangle with a angle being 31.0 degrees the opposite side from the angle is Vy the adjacent side is Vx the hypothesis is 11.0m/s then i drew a chart with X and Y X side of the chart I have Vx= 9.428840308 a=0 then on the y side i have
    Vy=5.665418824 a=-9.8 the Vf=0m/s I used the formula Vf=Vi+At and solved for time and got
    0.578103962 then i used d=Vi*t+1/2At^2 and solved for D and got 1.637620735 for the first question (But was told this was not correct)

    For the second question i doubled the time and multiplied by Vx D=9.428840308 (1.15620784) and got 10.90169909 but it was wrong !

    Is there anyone who might be able to help me Please??? :cry:
  2. jcsd
  3. Oct 31, 2006 #2
    what is the final velocity in the Y direction? And why?

    for your second question how do you maximize the distance you could travel? They gave you a hint : the SPEED is thesame but something else must be different. What must change?
    and Why did you double time??
  4. Oct 31, 2006 #3


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    GOod grief you don't believe in rounding numbers do you????

    If you've calculated, as it appears, the time taken for the vertical velocity to reach 0M/s, then you've calculated the time for the stone to reach it's maximum height, - it still has some way to go before it hits the ground.

    I would use here the equation x = V(i)t + 0.5 a t^2
    with the vertical velocity and a = 9.8 but remembering that when the stone hits the ground its vertical displacement is zero because it started from ground level

    That will give you a time, then use this with the horizontal velocity (constant) to calculate the horizontal distance.
  5. Oct 31, 2006 #4
    final velocity in y direction is 0 because at the apex it is not going anywhere
  6. Oct 31, 2006 #5


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    But the apex is not the final velocity. It comes back down again.
  7. Oct 31, 2006 #6
    tahts fine

    but is that the final velocity in the y direction for the WHOLE trip

    after all you are trying to find the time it took for the ENTIRE journey
  8. Oct 31, 2006 #7
    I get x= 1.637600545 is this right? (i am too stressed to round any numbers LOL)
  9. Oct 31, 2006 #8
    howd you get it?
  10. Oct 31, 2006 #9


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    That's not what I got.

    What to you get for the time?
  11. Oct 31, 2006 #10
    i took 5.665418824(0.578103962)+0.5(-9.8)(0.57810396)^2 and got 1.637600545
  12. Oct 31, 2006 #11


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    Homework Helper

    Did you use [tex]y(t) = 11\cdot\sin(31)t-\frac{1}{2}gt^2=0[/tex] to calculate the time?

    If you didn't, do so. (Solve for t.) When you do so, plug that time into [tex]x(t)=11\cdot\cos(31)t[/tex] to get the total displacement.
  13. Oct 31, 2006 #12
    to get the time i used the formula Vf=Vi+at
    a= -9.8
    Vi=5.665418824 (by 11.0sin(31.0))
    and then i solved for time
  14. Oct 31, 2006 #13
    why are you still saying velocity final in the Y is zero? Its NOT zero

    after it has completed its entire flight what is the velocity in the Y just before it hits the ground? I certainly hope it isnt zero otherwise whatever ive been doing for these 6 odds years has been in vain! VAIN!

    and this world has no justice :frown:
  15. Oct 31, 2006 #14
    I don't understand? I am going by the example we did in class and for Vf we used 0 and this is at the apex
  16. Oct 31, 2006 #15


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    Interesting class, indeed.

    Btw, I suggest you do a few searches named 'projectile motion', this could do good. :smile:
  17. Oct 31, 2006 #16


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    Staff Emeritus
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    Gold Member

    That would be good to calculate the maximum HEIGHT of the object, not the maximum distance it travels.

    Throw a ball into the air. When it hits the ground, is the y velocity zero? Use intuition, it really helps sometimes
  18. Oct 31, 2006 #17
    since im such a nice guy i drew a diagram for you
    look at the diagram

    now tell me what do u think the velocity i nteh Y direction must be after thw whole trip?

    THINK before you answer

    but dont think too hard :wink:

    you are probably mixing up the type of questions

    Attached Files:

    • 1.JPG
      File size:
      10.6 KB
  19. Oct 31, 2006 #18
    yes you are a nice guy but not only am i having trouble with physics BUT I CAN'T OPEN THE ATTACHMENT THIS IS SO SAD!!!!!!!!
  20. Oct 31, 2006 #19


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    Homework Helper

    ..since I'm an even nicer guy, I'll give you this Java applet to play with, and hopefully learn something from it: http://www.walter-fendt.de/ph11e/projectile.htm" [Broken].

    Last edited by a moderator: May 2, 2017
  21. Oct 31, 2006 #20
    correction SHE oooops :biggrin:

    k does the image work now?? look at the image. What is the final velocity in teh Y answer that question. THINK before you answer it

    And tell me WHY WHY WHYYYYYY you said that
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