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Shotput problem involving projectile motion

  1. Nov 29, 2007 #1
    So I reasoned the key to the first part is to find [tex]v_0[/tex] so that I can use it to find the force with which the shotput was thrown. I then divide that force by g to find the mass that the shotputter can lift.

    I started by finding the total time the shotput was airborne. I solved the following equation for t quadratically:

    [tex]y = -\frac{1}{2}gt^2 + v_0 sin\theta t + y_0 = 0[/tex]

    I found [tex]t = \frac{v_0 sin\theta + \sqrt{v^2_0 sin^2\theta + 4}}{g}[/tex]. Plugging this value into the equation [tex]x = v_0 cos\theta t = 22[/tex] I chugged through and found [tex]v_0 = 14.6 m/s[/tex].

    Now I go back to the first part of the throw, before the shotput is released. Using [tex]v^2 - v_0^2 = 2 a (x - x_0)[/tex] I can arrive at:

    Weight shotputter can lift with one arm = [tex] \frac{F}{g} = \frac{ma}{g} = \frac{mv^2}{2gx}[/tex] and I find the mass to be 39.4 kg. The book lists 42.0 kg as the answer so I'm not sure where the discrepancy lies since I didn't round until the very end when I did the problem.

    The second part is beyond me. If I take the range equation I used above, namely,

    [tex]x = v_0 cos\theta t = \frac{v_0^2}{2g} sin2\theta + \frac{\sqrt{v^2_0 sin^2\theta + 4}}{g}[/tex]

    the derivative is horrendous and I'm not sure I can solve it for [tex]\theta[/tex]. Also, the little formula the back of the book has indicates that there must be some other, more tractable, range formula to differentiate. I can't seem to figure out where to start in finding an alternate way of doing it. This also makes me think that there might have been a cleaner way of doing the first part. Am I right in thinking that or am I just missing something obvious here?
     
  2. jcsd
  3. Nov 29, 2007 #2
    Well, for part two, I would think that 45° is the most effective angle.

    Oh - here's your problem, I think - V_0sintheta (too lazy to do the latex) is not 22. 22. That is (probably) the distance from the ground at the putters foot to the final position. The actual distance would be slightly shorter, wouldn't it?
     
  4. Nov 30, 2007 #3
    The book is listing an angle that is slightly smaller than 45 degrees as the optimum angle.

    I agree that it should be less than 22 m in reality, but there's no way to figure it out from the problem. Just am I'm ignoring air resistance I should also assume the thrower is infinitely thin, I suppose.
     
  5. Nov 30, 2007 #4
    Well, I was referring to the fact that the distance is measured from the ground, whereas the actual is 2m up.

    I've recently conducted a lab nearly identical to this, but I can't for the life me remember how. :cry:
     
  6. Dec 3, 2007 #5
    Okay, I came back to this problem and managed to solve the first part (it was a mixture of a careless mistake and a book typo).

    I've come up with the following relation to solve part a:

    [tex]R sin2\theta + d(1 + cos\theta) = \frac{R^2}{R_0}[/tex]

    where d is the drop (2 m), R is the range, and [tex]R_0 \equiv \frac{v_0^2}{g}[/tex].

    I still have no clue how to find the optimum angle, however. If I solve for R and find the derivative it's a mess with thetas in each term. For this problem the book has the angle as:

    I have no idea how to derive that term because my derivative doesn't look like it can reduce to such a simple equality. Any ideas where I can start?

    The book gives a hint asking to show that the condition for max range is [tex]tan2\theta_m = \frac{R_m}{d}[/tex]. Obviously I'm not sure where to start on this one either. :cry:
     
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