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Projectile Motion: correct answer, but still confused

  1. Aug 20, 2008 #1
    1. The problem statement, all variables and given/known data

    This is Chapter 4 problem 45 of Halliday/Resnick/Walker Fundamentals of Physics 8th Edition. I can get the right answer (I'll show you...) but when I apply this answer to the problem I get a weird result for another derived value in the problem: the horizontal displacement of the projectile at the time it hits the target. I am probably missing something obvious or maybe subtle.

    Problem

    The Archer fish is able to shoot drops of water at insects sitting on branches near the water, and knock them off and eat them. A particular fish is at a stright-line distance of .9m from an insect. The angle from the surface of the water up to the line of sight to the insect is 36 deg. The fish shoots the drop at a velocity (Vo) of 3.56m/s. The task is to find the launch angle the fish should shoot the drop to hit the insect, subject to the constrain that the drop should be at the top of its parabolic path when it hits the insect.



    2. Relevant equations
    These are described below.


    3. The attempt at a solution


    The distance from the fish to the insect is .9m and the angle up to the insect is 36 degrees, so we have a right triangle with a hypotenuse of .9m and one angle of 36 deg. So to get the height of the insect above the water (h) we know sin(36)=h/.9m, so h=.53m. We can also get the horizontal distance (x) from the fish to the point under the insect since cos(36)=x/.9m, so x=.73m.

    To get the launch angle to shoot the drop (B) we use two equations.

    The first is the projectile motion equation for the height y: y=Vo sin(B)t-(g/2)t^2, with Vo=initial velocity, B the launch angle, t the time and g the free-fall acceleration (9.8m/s^2).

    The second is the equation for the y-component of velocity Vy=Vo sin(B)- gt.

    We know that at the top of its parabolic path Vy=0,and the problem says this is when the drop hits the fish. So by the second equation Vo sin(B)-gt=0. So t=Vo sin(B)/g. We plug this expression for t into the first equation for the height and after simplifying get y= Vo^2 (sin(B))^2/(2g). We know y is .53 by the geometry of the problem. So we can solve this equation for B and get B=arcsin ( (2g y)/Vo^2)^(1/2). When you do this out you get B=64.87 degrees. This is the answer in the book.

    We can get the time to hit the insect by using t=Vo sin(B)/g, i.e. t=(3.56m/s)(sin(64.87)/(9.8m/s^2)= .33 seconds.

    We can check this solution by plugging in this t=.33s and B=64.87 deg into y=Vo sin(B)t-(g/2)t^2 and get .53m which is the height of the insect by the geometry of the problem.

    NOW HERE IS WHERE I GET CONFUSED:

    If the launch angle is 64.87 deg and the initial velocity is given as 3.56m/s (which we used to correctly solve the problem) then the component of the initial velocity in the x-direction is 3.56 cos (64.87)=1.51m/s. Since this is projectile motion this x-component of the initial velocity (call it Vx) does not change with time. So we should be able to use this Vx and the time we hit the drop (.33s) to get the horizontal distance from the fish to the point underneath the drop. So we just do velocity times time: (1.51m/s)(.33s)=.49m. But the geometry of the problem showed us that the horizontal distance from the fish to the point under the insect is .73m, not .49m. So the projectile motion equations give the correct y but not the correct x for the point where the drop hits the insect.

    I'm probably missing something obvious...any help appreciated.
     
  2. jcsd
  3. Aug 20, 2008 #2
    From what I can tell, you are correct in that the problem might be bugged. Having a known delta-x distance and delta-y distance, along with two kinematic equations (in x and in y) leads to two equations in two unknowns (time, angle) that can be solved for time and angle. The requirement for the water to hit the bug at the top of it's trajectory leads to a third equation which appears to be inconsistent with the first two. In this case the problem would have no solution.
     
  4. Aug 21, 2008 #3
    The problem is bugged. The derivative of the parabola intersecting (0,.53) and (-.73, 0) is 1.452 at (-.73, 0). Arctan(1.452/1)=55.45 degrees. Therefore, with x=.73 and Ymax at .53 the geometry tells us that the projectile would have to be launched at 55.45 degrees to intersect (0, .53) at the top of its parabolic path.

    To further prove this, eliminate Vo^2 from the range equation and from the Ymax equation and then solve the parametric equation for the angle (R=2(.73) and Ymax=.53) This should spit out the same angle as the above calculus exercise (55.45 degrees). To find the velocity at which the projectile would need to be launched given the above values, solve for Vo in both the range equation and the Ymax equation and make sure they agree, Vo=3.92m/s.

    The bug in the problem is the fact that it is impossible for the projectile to be at the top of it's parabolic path when launched at 3.56m/s when x=.73, but because the Ymax equation doesn't depend on x, it gave you a correct angle with Vo given and Ymax given.

    You keenly discovered that x had to be .49m instead of .73m using a basic kinematic equation. You could have also discovered it by eliminating Vo^2 from the Range equation and the Ymax equation then solving for the Range and then dividing it by 2. This gives .497m.
     
  5. Aug 21, 2008 #4
    Thank you for your two very thoughtful replies. Just so I'm clear, the bug is that the problem specifies a max height and an initial velocity, and a constraint of Vy=0 at impact, and these don't give the x-displacement that is specified in the problem.

    I did the math of eliminating Vo and got the 55.46 angle for .73m displacement. This angle then gives a Vo of 3.91m/s.

    Glad I wasn't missing something obvious...Thanks again.
     
  6. Aug 21, 2008 #5
    Haha, bugged. I get it. If you're doing this through some online homework program it's not unlikely it was coded wrong. Sometimes they randomize numbers they shouldn't, or vise versa. But I read somewhere the archer fish waits until it's directly underneath the insect, otherwise it would miss because the light gets all bent between the air and water. It eventually learns to adapt, but not very well cause of the tiny brain and all. Try that answer, see if you get extra credit
     
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