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Homework Help: Honors physics basic projectile motion question

  1. Jun 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Question: Inbar and Luc are captured by angry natives on a desert island, after their plane crashes. They need to jump over a pit filled with snakes. They jump up at an angle of 23 degrees and need to land 8.7 meters away in x. What does their take off speed have to be?

    The variables given are in the problem. My biggest problem is I don't understand how to combine the x and y equations in projectile motion. Here are the variables I think I have but please correct me if I am wrong.
    Is there anything else I need to find or any other variables I should already know from the problem?

    2. Relevant equations
    Here are some equations which I have tried using.
    X=Vi,avg t

    3. The attempt at a solution
    So here is what I have tried. It's not much but it's my small attempt.
    First I set up a right triangle trig problem. Then I used tangent to find that viy=3.7. Next I used Vfy=Viy+ayt to try to find t. From here I get kinda stuck. I don't even know if the first part was correct but at least I tried.
  2. jcsd
  3. Jun 18, 2011 #2
    So this problem requires a little bit of abstract thought.

    We know that the x velocity, Vx has to be just enough to clear the snakes, so 8.7/t, where t is your hang time.

    By the trig of the problem, we can also say that Vx = V cos(23).

    We know that Vy has to be just enough so that produces the desired hang time, t, which equals 8.7/Vx = 8.7/(V(cos(23)).

    So we have:

    t = 8.7/(Vcos(23))

    So need to figure out the y component.

    We can use the equation Vf = Vo + at and apply it to the Y kinematics.

    So -Vy = Vy - 9.81 * t (We say -Vy is Vf because when he lands, his Y velocity will be opposite but equal his up Y velocity).

    From here, we can say Vy = Vsin(23).

    We now have our two equations.

    t = 8.7/(Vcos(23))
    -Vsin(23) = Vsin(23) - 9.81 * t.

    Two equations, two unknowns. Happy Hunting.
  4. Jun 18, 2011 #3
    I am a bit confused about the V in Vsin(23) and Vcos(23). Is the V in x or Y? Is it Vi or something like that?
  5. Jun 18, 2011 #4
    V is the final velocity.

    If you draw your triangle, you have V going off at an angle, and Vx horizontal and Vy vertical. So you can say that Vx = V * cos(23) while Vy = V * sin(23).
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