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Homework Help: Projectile Motion Problem

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data

    At a time t, a projectile is measured to have a height Y = 12 m above the ground and a velocity v = 4 m/s at an angle (theta) = 42 with respect to to the horizontal. At what horizontal positions is the particle at a height 3/4 Y ?

    Height= 12 m
    Velocity= 4 m/s
    Angle= 42 degrees

    2. Relevant equations

    I believe both the range equation x= x0 + V0xt and y= y0 + V0yt - (1/2)gt2

    3. The attempt at a solution

    I tried to to use the second equation, but I had trouble with time since it was not given. I used 4sin(42) for V0y, but I don't think it helped me much.

    This should be simple, I can tell, but I am just stuck.
  2. jcsd
  3. Jan 30, 2013 #2
    You are on the right track with your attempt but you need to find a relationship between the two equations. (HINT: What do the x position and Y position equations have in common)
  4. Jan 30, 2013 #3
    They both have initial velocities, and those can be figured out using the velocity given and the angle. X0 would be 4cos(42) and Y0 would be 4sin(42).

    Am I getting any closer?
  5. Jan 30, 2013 #4
    Think variable. Both functions are dependant on time.
  6. Jan 30, 2013 #5
    Would I have to set them equal to each other by solving for time? I know that the time is the same for both directions.
  7. Jan 30, 2013 #6
    I also recently tried using the quadratic formula with my Y equation, but I end up getting a negative under the radical. I am not sure why.
  8. Jan 30, 2013 #7
    Remember that the generic quadratic eq is ax^2+bx+c. You should get a positive number
  9. Jan 30, 2013 #8
    If I use y= y0 + V0yt - (0.5)gt2

    I get

    12= 0 + 4sin(42)t-(9.8/2)t2

    In the general equation form, it is:


    Unless this is incorrect, my answer is negative under the radical.
  10. Jan 30, 2013 #9
    y_0 is 12 not zero. Y should be 3/4
  11. Jan 30, 2013 #10
    So what would be the final Y position? It seems like since it is in motion, it would start at zero.
  12. Jan 30, 2013 #11
    From the problem we can assume that at t=0 the Y position is 12m. The last line ask us to solve from is 3/4 of Y or 0.75*12. Thus your equation for Y should look as follows:
    0.75 *12 = 12 + 4* sin(42)*t - 9.8/2*t^2
  13. Jan 30, 2013 #12
    Okay. Yes, that looks like what I have after fixing it. I must be creating some sort of error in my calculations then, because I plugged the equation : -4.9t2 + 4sin(42)*t +3 into a graphing calculator, and there is indeed a positive value.

    It is odd. Using the quadratic formula, underneath the radical, which is sqrt(b^2-4ac),
    b^2= (4sin(42))^2, but 4ac is a bigger number, being : 4 (-4.9)(3)= 58.8.

    I might just be doing something wrong I suppose.
  14. Jan 31, 2013 #13
    You are forgetting you negative sign. A = -9.8
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