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Totally Lost (Free Fall Distance)

  1. Sep 8, 2005 #1
    Hi, its me...AGAIN!!! You people should start CHARGING for your services!

    I have a conceptual problem more than just a homework problem this time. It regards free fall distance and the formula applied.

    Okey dokey...the formula to find out how far an object falls from rest is:
    d=1/2 gt^2

    So I did some sample problems to get a hang of it...the problems consist of shooting an arrow straight up at 50m/s.

    One problem is: Where is the arrow you shoot up at 50 m/s when it runs out of speed? Answer: 125m. I get this by multiplying 50(g) by 5(t) and dividing by 2.

    Next problem: How high will the arrow be 7 seconds after being shot up at 50m/s? Answer: 105m. I got this by starting with the above answer, 125m because that is the distance traveled until the arrow stopped. To account for the 2 extra seconds, the arrow is falling at 10m/s so I subtracted 20 from 125 to get 105m.

    My trouble is coming from the fact that I never used the ^2 portion of the formula to come to these conclusions. I simply multiplied acceleration and time and divided by 2. When was I supposed to square the time, and why did I come up with the correct answers without doing it???
     
  2. jcsd
  3. Sep 8, 2005 #2
    You don't seem to understand the equation or concept.
    The basic kinematic equation: [tex]x=v_0t+\frac{1}{2}at^2[/tex]
    The acceleration in this case is g which is [tex]-9.81 \frac{m}{s^2}[/tex]
     
  4. Sep 8, 2005 #3
    ...One problem is: Where is the arrow you shoot up at 50 m/s when it runs out of speed? Answer: 125m. I get this by multiplying 50(g) by 5(t) and dividing by 2.

    >> the constant 50 must not involve in the solution to get 125m and you use g= 10m/s^2 rather than 9.8m/s^2..
    ________________________________________________________________________
    ...My trouble is coming from the fact that I never used the ^2 portion of the formula to come to these conclusions. I simply multiplied acceleration and time and divided by 2. When was I supposed to square the time, and why did I come up with the correct answers without doing it???


    >> it really return the same answer but not at all time, its just a coincidence
    the exact solution is from the formula; y= 1/2g*t^2 (no initial velocity from the point the ball out of speed at t=5s)

    you use g=10m/s^2

    1. right: y= 1/2*10*2^2= 20m; 125-20= 105

    2. your answer: y= 1/2*10*2= 20m; 125-20= 105

    reason: it is just a mere coincidence that the time we encounter here is 2s, squaring the quantity of 2 would result to 4 and it is just like when you multiply it with 2 (2^2=4; 2x2=4). However for example, if we use t= 3s then it surely return different answer because squaring 3 is 9, but multiplying 3 by 2 it is 6..Now for the sake of consistency with the units we must not ignore squaring the quantity of time because if we will ignore it and just multiply it with 2 the resulting unit will be "m/s" and not "m".
     
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