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Plot the range r

  1. Sep 21, 2014 #1
    We are given the velocity at which an object is projected (30 m/s) at an angle theta. I have to "Plot the range R as a function of angle theta (a value somewhere between 0 and 90 (including 0 and 90)). My thought was to possibly determine R using theta = 0 and theta = 90 to find a range, but I'm not sure that's right. The next question then asks "For what angle theta is the range R maximum?" How would I even find this?
  2. jcsd
  3. Sep 21, 2014 #2
    For any particular value of ##\theta##, you have a particular range ##R##. So you have a function: ##R(\theta)##. You need to find a formula for this function and then do what you are asked.
  4. Sep 21, 2014 #3


    Staff: Mentor

    Do you know how to find the maximum of a function?
  5. Sep 21, 2014 #4
    No and I've read the chapter in my physics book twice, but still don't understand
  6. Sep 21, 2014 #5
    If the problem said, "find the range of a projectile launched at angle 21 degree to the horizontal with the initial velocity 30 m/s", how would you go about it?
  7. Sep 21, 2014 #6
    R(θ) = Vo2 sin(2θ)/g

    R = (30^2 * sin(21))/9.8 = 32.91 m
  8. Sep 21, 2014 #7
    And now if the problem said the same, except "at angle 37 degree", I am sure you would be able to find the answer much in the same way, which is good. If you repeat that for a few different values of the angle, you will be able to plot the function. Just make sure that you cover the entire range 0 - 90 with values not too far away from each other. Does the plot suggest where the maximum might be?
  9. Sep 21, 2014 #8
    I solved using theta values = 0, 10, 20, and so on. It increased from R=0 to R=91.83 at theta=90. This was the highest value, but am I doing something wrong? If you shoot an object straight into the air, shouldn't it come straight down?
  10. Sep 21, 2014 #9


    Staff: Mentor

    That is usually a topic that is covered in a math class. The way you do it is to take the derivative of the function and set it to 0. So here you would take ##dR/d\theta=0## and solve for ##\theta##
  11. Sep 21, 2014 #10
    Your formula is correct. However, the results you are getting are not, because ## \sin 2 \cdot 90^{\circ} = \sin 180^{\circ} = 0 ##. Check your calculations, and make sure your calculator is in the degrees mode.
  12. Sep 21, 2014 #11
    Okay!! I forgot to multiply all of the numbers by 2! I'll do that now. Thank you so much for your help! You don't even understand how much it means to me!
  13. Sep 21, 2014 #12
    I'm in an algebra based calculus class. He doesn't want us using calculus. That's probably why we weren't taught the derivative part.
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