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Just can't figure this one out

  1. Jul 21, 2004 #1
    Please bear with my verbosity on this one, and with the messy symbols. :yuck:

    The problem states that I am to fire a projectile from some distance d above the ground (ignoring air resistance) with initial velocity v0 and initial angle θ. I am to show that between R (the maximum horizontal distance), v0, θ, and the drop d there is a relationship R sin2θ + d (1 + cos2θ) = R²/R0 where R0 ≡ v0²/g.

    Well, the way I see it the range has two parts. The first part is the range R1, the horizontal distance the projectile covers in going from y=d to some ymax and then falling back to y=d. This is given by the normal
    (v0² sin2θ)/g. Then one should have to calculate the horizontal distance covered by the drop from y=d to y=0, which is simply R2=v cosθ t', where t' is the time it takes for the ball to drop from y=d to y=0. Well, at y=d the object should have velocity v0 and be falling at angle θ. To find t' I use the formula y=y0 + v0 sinθ t' - ½ g t'². Plugging in y=0 and y0=d and using the quadratic formula I get that:

    t' = [v0 sinθ ± √(v0² sin²θ + 2gd)]/g.

    Now, adding R1 and R2 to get R, I obtain:

    R = R1+R2 = (v0² sin2θ)/g + (v0² sin2θ)/2g - [v0 cosθ √(v0² sin²θ + 2gd)]/g

    or R = (3 v0² sin2θ)/2g - [v0 cosθ √(v0² sin²θ + 2gd)]/g

    Well, I can't manipulate that equation to get the necessary relationship, and in a related problem utilizing the above equation didn't give me the correct answer (they asked for an maximizing angle and a v0).

    The logic seems pretty sound to me, so what am I doing wrong? I've been agonizing over this for hours, so any help is appreciated.
     
    Last edited: Jul 22, 2004
  2. jcsd
  3. Jul 22, 2004 #2

    arildno

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    Welcome to PF!
    One obvious mistake you've done is in your calculation of R2:
    "at y=d the object should have velocity v0 and be falling at angle θ. To find t' I use the formula y=y0 + v0 sinθ t' - ½ g t'²."

    The projectile is FALLING, not RISING!!
    Hence, surely, you must use:
    y=y0-v0sin(theta)t'-1/2gt'^2

    (I haven't checked the rest yet)

    Note:
    The formula you use is valid from the initial point, i.e, when the projectile is shot upwards.
     
    Last edited: Jul 22, 2004
  4. Jul 22, 2004 #3
    Thanks for the reply!

    I can't believe I made such a basic mistake. With that sign change, I'm able to manipulate the terms and obtain the final answer. Much appreciated.
     
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