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I have 2 physics problems, and I need help.

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data
    I've been absent for quite a bit in school and I have a lot of homework and catching up to do. I've been looking for many resources at hand which can help me understand these problems but so far the book is not helping me. I missed some of the vector lessons, a lot of the friction lessons and we are already into Projectiles launched at an angle.
    So here are my problems:

    1) In a scene in an action movie, a stuntman jumps from the top of one bulding to the top of another buliding 4.0 meters away. After a running start, he leaps at an angle of 15 degrees with respect to the flat roof while traveling at a speed of 5.0 m/s. Will he make it to the other roof, which is 2.5m shorter than the building he jumps from?

    2) Salmon often jump waterfalls to reach their breeding grounds. Starting 2.00 m from a waterfall 0.55 m in height, at what minimum speed must a salmon jumping at an angle of 32.0 degrees leave the water to continue upstream?

    2. Relevant equations
    1) I have been using the following equations to solve the answer but I'm not even sure what I am looking for :s.

    vf2 = vi2 + 2ad
    d = (vf2 - vi2) / 2a To see the maximum total vertical distance

    a = (vf - vi) / t
    t = (vf - vi) / a For the time he is in the air.

    2) Really have no idea what I even did.

    a^2 + b^2 = c^2.

    3. The attempt at a solution

    5(sin(15))= 1.295 m/s.
    5(cos(15))= 4.83 m/s.

    T= (0-1.295)/(-9.81) = 0.264 seconds.

    (1.295^2-0)/(2(-9.81))= 0.085 m = Dy.

    4.83m(0.264s) = 1.275198 m.
    Not sure what I even did, but it is some relevant work to this problem. I'm not sure what I am even looking for and if someone could please help it would be appreciated.

    I was really confused when I tried this problem, any help would be appreciated.
    Will edit if proof is needed for attempt.
  2. jcsd
  3. Nov 20, 2008 #2


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    I can give you a general strategy that often works in such "jumping" problems. The only thing you really need to remember is a general formula for motion and how to solve quadratic equations, and you can do virtually any exercise such as the two you posted.

    The key to solving such problems is to remember, that horizontal and vertical motion are independent. That means you can solve
    [tex]x(t) = \frac12 a_x t^2 + v_x(0) t + x_0[/tex]
    [tex]y(t) = \frac12 a_y t^2 + v_y(0) t + y_0[/tex]
    independently, where [itex]a[/itex] is the acceleration in the x and y-directions, [itex]v(0)[/itex] the initial velocity, and [itex](x_0, y_0)[/itex] the initial position.

    Probably on first read, that all sounds a bit abstract, so lets try to solve a problem with it. I'll talk you through it, but first I want you to identify the variables:
    • For the first problem, the initial velocity and the angle are given. Determine from those the horizontal and the vertical component of the velocity, [itex]v_x(0), v_y(0)[/itex].
    • What is the horizontal acceleration? What is the vertical acceleration? What is the initial horizontal position, and the initial vertical position?
  4. Nov 20, 2008 #3
    For number 1 this is what I'm getting:
    The horizontal component is 4.8296 m/s, and the vertical component is 1.294095 m/s.
    The horizontal acceleration 0, I'm not sure of this. I thought there is no change in the horizontal component of an object's velocity.
    The vertical acceleration is gravity? -9.81m/s/s.

    The initial horizontal position is 0, 4 meters away from the other building. Or maybe -4?
    The initial Vertical position is 2.5 m higher than the other building, so 2.5m.

    Also I really appreciate the help you are giving thanks.
  5. Nov 20, 2008 #4


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    Yep, it's just 5 cos(15 degrees) and 5 sin(15 degrees).

    Very good, most students tend to get confused. As I said, the key is to treat horizontal and vertical separately and indeed the horizontal component does not change.

    Yep, how you do it is up to you. You can choose it whichever way you want, as long as you explicitly write it down. So you can say, the x-coordinate of the starting point is 0, and then that of the end point is 4. Or you can take -4 for the start and 0 for the end point. Or you can take -2 for the start point and 2 for the end point... although that's a bit strange choice :smile:

    Don't thank me yet, we're not through :smile:

    In summary then, we have (using x0 = 0, y0 = 2.5)
    [tex]x(t) = 1.294\ldots t[/tex]
    [tex]y(t) = -\frac12 \times 9.81 t^2 + 4.82\ldots t + 2.5[/tex]
    (watch the signs: the initial velocity is + and the acceleration is -)

    Now you can solve, for example, how long it takes to reach y = 0 from the second equation, plug that into the first, and see if he makes x = 4. Or, you can solve from the first when he reaches x = 4 (easier) and then plug it into the second if y >= 0 at that moment.
    I trust you can do the math?
  6. Nov 20, 2008 #5
    I'm sorry, but can you explain the equations you gave. I'm trying to understand this section as much as possible and my lack of knowledge of all of the equations and what some variables represent gets in my way.

    I'm not following what the dots represent :S, if you can, please elaborate.

    I'll try to do what I can as of now and see what I can come up with.

    [tex]4m = 1.294 \times t[/tex]

    [tex] t = 3.0911 seconds[/tex]

    [tex]y(t) = \frac{-9.81(3.09)^2}{-2} + 4.82(3.09) + 2.5[/tex]

    [tex]y(t) = -29.43 [/tex]

    That can't be right lol, I must've done something wrong.
  7. Nov 20, 2008 #6


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    Sorry, the dots are just there to indicate that the values 1.294 and 4.82 were obtained from another calculation (namely 5*sin(15 degrees) and 5*cos(15 degrees)) and that you should take care when using them not to round off too much (or, preferably, not at all). Instead of writing down 5 decimals, I prefer just to write 2 or 3 and put some dots to indicate that I am not rounding off in the intermediate step (on my TI calculator, for example, I would not type 4.82 but use the Ans key).

    For the rest, I just put in the numbers you gave me. In the first formula, a = 0 and x(0) = 0, so you only get the part with v(0) t, and you had calculated v(0) already. In the second one, I used a = -9.81, so the first term is [itex]\frac12 \times (-9.81) t^2[/itex] and then I put the minus sign in the front: [itex]- \frac{9.81 t^2}{2}[/itex]. You smuggled in an extra minus sign, and your answer became a bit strange (as you correctly noticed).

    The calculation looks pretty all right though, apart from that sign :smile:
  8. Nov 20, 2008 #7
    Thank you so much, I'll try to do the second one myself.
    Be sure to check it if you can, I'll update my work for that one as well.

    One thing I don't understand is what does the answer represent, is there more to the problem? Do I have to use the equations you stated in the first post you gave?
    I checked the back of the book and the answer is as follows:

    [tex] \Delta Y = -2.3 m [/tex]
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