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Trajcetory of a projectile

  1. Mar 18, 2014 #1
    1. The problem statement, all variables and given/known data

    The question is about the darts game. Bulls eye is at 1.73m height. The Dart (m=25.3g) is thrown from a distance of 1.80m.
    Dart lands in the middle of the board (bulls eye) under an angle of 4° (to the horizontal line) with landing speed of 12m/sec. Air resistance is to be ignored.

    Things I have to know are:
    1.) How long was the dart in air?
    2.) At what height (h0) was the dart in the moment of throwing?
    3.) What was the angle of the direction the dart was thrown at in the moment of throwing (again, to the horizontal line)
    4.) What speed was the dart thrown at?
    5.) What was the highest point of the dart flight line?

    2. Relevant equations

    3. The attempt at a solution

    1.) Since vx doesn't change, as far as I know: 1.80m / 12m/s = 0.15s.

    2.) Now, I tried using different equations and even drawing this, but I always seem to be missing at least one piece of information, usually the throwing angle. For example using y=(v0*sin[itex]\alpha[/itex]0)*t-0.5*g*t2.. I tried to calculate v0 by separating the vector (speed and angle) at the landing in y and x vectors. Then used the vy vector (sin(4)*12=0.837m/s) and set it up in this equation vy=9.81*t+C, getting -0.6345 for C. Found out that vy=0 in t=0.065sec and calculated the positive and negative surface, added them and got 0.015m. I thought this would be the height difference h0 - h, but Maple (where I have to put my results in) says that's incorrect.

    3.) Here, once again, I always felt like I was missing some information, no matter which formula I chose. Couldn't really think of anything else I could use to somehow get a result here..

    4.) I used vx=v0*cos([itex]\alpha[/itex]0) here and changed it a bit and got 12m/s / cos(4) = v0 = 12.029m/s. Which is correct, according to Maple.

    5.) In case what I did in 2.) was correct, the highest point should be in t=0.065 sec. I believe that could be calculated using h=h0-v0*t-0.5*g*t2. But since I'm missing h0, this doesn't help much.
  2. jcsd
  3. Mar 18, 2014 #2

    Might I suggest working backwards in this problem?

    Also remember that your landing speed gives you two values: your velocity in the x direction and your velocity in the y direction, all you need is to use trig to find those two components.

    From there, you are correct in dividing your distance by the velocity, but you used the speed and not the x component of the velocity (although it is close) to determine the total time your dart was in the air.

    I suggest using your y component of velocity to find out how long it would take to fall from the vertex of the trajectory (where your initial velocity in the y direction would be zero!) to your final landing velocity in the y direction.

    Once you find that time, you have pretty much everything you need! The rest is plugging into kinematics!
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