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Three Dimensional Projectile motion

  1. Sep 22, 2016 #1
    1. The problem statement, all variables and given/known data
    A baseball player throws a ball at an initial velocity of 27m/s. It lands 140m away at 6.4 seconds.
    (A) what is the angle he threw it at?
    (B) what is the maximum height?

    2. Relevant equations


    3. The attempt at a solution
    I found the angle to be 36 degrees using the formula d=ViΔt+1/2aΔt^2
    140 = 27cosθ(6.4)
    140/6.4 = 27cosθ
    θ=36 degrees


    However when I went to calculate the maximum height I I subbed it into my formula:
    dxmax= 27sin36•3.2+1/2(-9.81)•3.2^2
    dxmax=0.5574m

    Which I find quite unreasonable?
     
  2. jcsd
  3. Sep 22, 2016 #2

    kuruman

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    This is the wrong way to go about it. You are given a range of 140 m, but as you know there are normally two projection angles that give the same range. How do you know you got the correct one? Can you think of another approach to the problem? I am glad you found the answer unreasonable, because it is.

    On edit:
    Actually, this is a poorly constructed problem. If you accept the range of 140 m and the time of flight of 6.4 s and totally ignore the initial speed of 27 m/s, you get a reasonable answer, all quantities are defined and initial speed is not 27 m/s. That's because you can get the components of the initial velocity from the range and the time of flight, which is how I approached the problem.
     
    Last edited: Sep 22, 2016
  4. Sep 22, 2016 #3

    rude man

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    The problem statement gives more data than the problem needs, so either one of the data is wrong or it's redundant.
    In this case there are two results for the angle, one determined by the horizontal component of flight and one by the vertical.
    I hope, for the sake of U.S. physics teaching in general, the problem was misparaphrased, but it doesn't look like it with all those numbers.
     
  5. Sep 22, 2016 #4

    haruspex

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    But they would not take the same time, so there is no ambiguity in the angle.

    You cannot find the angle without that information.

    No, there's no redundancy. do not assume it was thrown from ground level (itcannot have been).
     
  6. Sep 22, 2016 #5

    kuruman

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    I completely missed this point and I thank you for pointing it out. However, there is still a problem. I ran the numbers and assuming that I did everything correctly, I get that the ball must have been projected from a height of 99.4 m. This is unrealistic for a baseball player even if his team is the San Francisco Giants. :smile:
     
    Last edited: Sep 22, 2016
  7. Sep 22, 2016 #6
    I attempted to find the value of the angle with my horizontal measurements using cos, then after with my vertical measurements using sine. I figured it out to be 36 like stated up above with cos. But I couldnt figure out sine. I wasnt sure how exactly to approach the problem differently. The teacher suggested to find the angle using cos due to the fact it requires less rearranging when it comes to the formula. Thanks for help.
     
  8. Sep 22, 2016 #7
    Yes I got way different answers when calculating angle using horizontal and vertical measurements.
     
  9. Sep 22, 2016 #8

    kuruman

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    Here are two approaches to the problem
    Your original solution. Does not work because if you assume that the ball takes 3.2 to reach maximum height, then if you calculate where it is in 6.4 s using y(6.4) = 27sin36•6.4+1/2(-9.81)*6.42 you get - 99.4 m, i.e. the ball is not at ground level, y(6.4) ≠ 0. Now if the ball were launched at height +99.4 m (as haruspex indicated), it would work, but contradicts common sense.

    Ignore the given initial speed and calculate v0x and v0y using the range and time of flight in the kinematic equations. Does not work because you find that the initial speed is greater than 27 m/s which contradicts what is given. Of course, if you are to ignore one of the given quantities, there is nothing that says that you should pick the 27 m/s as opposed to the 140 m. As rude man observed, "The problem statement gives more data than the problem needs..." For example, you can find the components of the initial velocity components v0x = 140/6.4 m and v0y = sqrt(272-(140/6.4)2). But if you do this and calculate the range using the formula R = 2* v0x*v0y /g, you get 70 m, not 140 m.

    I know all this does not help you much if you have to turn in a homework solution, that's why I recommend that you ask your instructor so that he/she knows that all is not OK with this problem.
     
  10. Sep 22, 2016 #9

    rude man

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    There is one and only one value for θ satisfying all three data points, but that assumes that the ball is thrown from a height of 85.1m.
     
  11. Sep 23, 2016 #10

    haruspex

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    Yes, I had not got around to working the numbers, but I had nasty feeling it would turn out to be unreasonably high.
     
  12. Sep 23, 2016 #11
    Can I ask how you got this? Because it just does not work out for me.
     
  13. Sep 23, 2016 #12

    Charles Link

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    An initial speed of 27m/sec is about 88 ft./sec. which is 60 m.p.h. That is a reasonable speed for a high school player with a moderately good arm. The throw even at an angle of 45 degrees (which gives the maximum distance) will go about 75 m (without any air resistance, if my algebra and arithmetic are correct) which is about 240 ft. and far short of 140 m. Without any air resistance a major league outfielder with a good arm (v=100 mph) might get 140m=450 ft. on a throw. (With air resistance included, 350 ft. is about as far as the ones with the very best arms can throw.) The t=6.4 seconds is reasonable if the initial speed were v=100 m.ph.=45 m/sec. (v=27 m/sec is completely inconsistent with the other two numbers) .In any case, they only need to give two of the three quantities, time in air, distance traveled, and initial speed to find the other quantity and also find the angle that the ball was thrown.
     
    Last edited: Sep 23, 2016
  14. Sep 23, 2016 #13

    haruspex

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    Did you read post #4?
    You correctly calculated the horizontal velocity, and hence the angle and vertical velocity.
    If we assume it was thrown at height h with that vertical velocity and hit the ground after 6.4s, what is h?
     
  15. Sep 23, 2016 #14

    Charles Link

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    Generally, when they introduce a baseball player, the 5 ft. height at the shoulders or thereabouts is ignored and they are talking about a very practical problem (and they are likely a big fan of our national sport (U.S.A.) baseball and might even play the game themselves). I still play a lot of baseball (I'm 61), and the question often comes up, how far can someone throw or how fast can they throw? For this problem, it is assumed the throw occurs at ground level (as it always does in a baseball game), and if this is the case, there is redundancy.
     
    Last edited: Sep 23, 2016
  16. Sep 23, 2016 #15

    haruspex

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    Sure, but given that the numbers and the apparent overspecification presented make no sense with that assumption, it is appropriate to pursue the alternative. As it turns out, h would have to be tens of metres, so the question is still likely mangled, but we should exhaust that path first.
     
  17. Sep 24, 2016 #16

    kuruman

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    In the absence of air resistance, the position, as a function of time, of any projectile launched from initial position ## \vec {r}_0 ## with initial speed ## \vec{v}_0 ## is given by the vector sum$$ \vec{r}(t) = \vec {r}_0 + \vec{v}_0 t - \frac{1}{2}gt^2 \hat{y} $$This expression is valid as long as the acceleration is constant, i.e. as long as the projectile is in free fall. Normally, the origin is chosen so that ## \vec{r}_0 = 0 \hat{x} + y_0 \hat{y} ##.
    Let tf = the “time of flight”. Then, the equation simplifies to
    $$ \vec{r}(t_f) = y_0 \hat{y} + \vec{v}_0t_f - \frac{1}{2}gt^2_f \ \hat{y} = R \ \hat{x} $$ where R is the range, or horizontal distance traveled. For this problem, an appropriate vector diagram, drawn to scale, is shown below.
    ProjectileAnalysis.png
    I constructed the drawing as follows: I first drew the horizontal range R = 140 units. Starting at the tip of R, I raised the vertical segment (1/2) gtf2 = 201 units. At the other end of the vertical segment, I drew a circle of radius v0tf = 173 units. Starting at the origin, I raised another vertical segment y0 to its point of intersection with the circle. Note that this geometrical solution incorporates all the given quantities and shows quite clearly how they fit together. Sometimes a geometrical representation is clearer than equations of motion written in the horizontal and vertical direction.
    When constructing projectile motion problems, it might not be a bad idea to look at the vector geometry and make sure that the numbers fit and that the problem is not overdetermined. Furthermore, the geometry can become a heuristic tool to uncover hidden relations without using extensive algebra. For example, in the simpler case where the ball lands at the same level as the launching point (i.e. y0 = 0), the trapezoid becomes a right triangle. The Pythagorean theorem says $$ R = \sqrt{(v_0 t_f)^2-(1/4)(gt_f^2)^2}=t_f \sqrt{(v_0)^2-(1/4)(gt_f)^2} $$ Since it is also true that ## R = v_{0x}t_f ##, it follows that $$ v_{0x} = \sqrt{(v_0)^2-(1/4)(gt_f)^2} $$ Who would have thought it?
     
    Last edited: Sep 24, 2016
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