Projectile Motion Range (please Help)

  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. 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. 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. final velocity in y direction is 0 because at the apex it is not going anywhere
     
  6. But the apex is not the final velocity. It comes back down again.
     
  7. 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. I get x= 1.637600545 is this right? (i am too stressed to round any numbers LOL)
     
  9. howd you get it?
     
  10. That's not what I got.

    What to you get for the time?
     
  11. i took 5.665418824(0.578103962)+0.5(-9.8)(0.57810396)^2 and got 1.637600545
     
  12. radou

    radou 3,215
    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. to get the time i used the formula Vf=Vi+at
    Vf=0m/s
    a= -9.8
    Vi=5.665418824 (by 11.0sin(31.0))
    and then i solved for time
     
  14. 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. 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. radou

    radou 3,215
    Homework Helper

    Interesting class, indeed.

    Btw, I suggest you do a few searches named 'projectile motion', this could do good. :smile:
     
  17. Office_Shredder

    Office_Shredder 4,499
    Staff Emeritus
    Science Advisor
    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. 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
     

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  19. 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. radou

    radou 3,215
    Homework Helper

  21. 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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