1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Constant Acceleration Problem

  1. Jan 13, 2007 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown upward from the edge of a 20m building with velocity V. One second later a ball is dropped from the same building. With what velocity must the first ball be thrown so that both balls land at the same time.


    2. Relevant equations
    x(t)=x+vx+1/2(ax)^2

    I'm am using x1=the first ball
    I'm using x2=the second ball

    vx^2=2(ax)d^2



    3. The attempt at a solution

    So I solved for how long it will take x2 to fall the height of the building with the equation

    0=20-1/2(10m/s)*t^2
    so t=2s

    So, I guess since it takes 2 seconds for x2 to fall the distance of the building, plus the extra second, x1 must take 3 seconds to fall.

    So know I guess I need to find with what velocity the first ball must be thrown so that it takes three seconds to hit the ground.

    The way the professor explained in lecture was by using the equation that velocity squared is equal to twice the acceleration times the distance.

    so Vx^2=2(ax)d

    he also divided the times into the time it takes from the ball to go from the roof to the peak of the throw, then from the peak to the roof, and then from the roof to the ground.

    So I guess the time from the roof to it's maximum height is by the distance equation also. So for x1

    y(t)=vx(t)+1/2(-10m/s)t^2
    y(t)=vx(t)-5t^2

    and since vx^2=2(ax)d
    vx^2=2(-10)d
    vx^2=-20d

    I dunno, I guess I didn't really follow how my professor did this. Also, i'm having a hard time realizing what the difference between distance and position is. I know that if I can find the velocity upward and the velocity as the ball is passing the roof on the way down. I can find the time for the final distance down the building using that initial velocity. I dunno, I'm lost. Any help is appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 13, 2007 #2
    The first thing to understand is that in the equation of motion, x(t)=x+vx+1/2(ax)^2, x(t) stands for displacement along x, and not distance, ie, its value depends only on the position of the body and not the path taken by it to reach that position.

    Now, select an origin to measure all x from ( it seems you have chosen the ground as origin ) and don't forget to use appropropriate sign conventions.

    Very correct. Now how do you go about finding v ?
     
  4. Jan 13, 2007 #3
    So, to find v I know that I am gonna have to split the total fall time of x1 into the time up and the time down. So, I know that I can get distance in terms of velocity for the way up.



    x1(t)=20+V0x*t+(1/2)*(-10)*t^2
    x1(t)=20+V0x*t-5t^2

    I know at the peak that velocity equals zero

    since Vx^2=V0x^2+2*(ax)*(x)
    0=V0x^2-5x
    5x=V0x^2

    So since I have three variables, and I also need to somehow solve for time equal to three seconds, I'm pretty stuck.


    So, I guess I need to find the value of distance in terms of velocity since I have two variables. This is where I'm running into the problem of distance vs. position.
     
  5. Jan 13, 2007 #4
    Of course, you can analyse the problem that way if you like, but a better way would be to consider the displacement. Displacement doesn't depend upon whether the body went up and then down or any other path. It is only concerned with the position of the body at any instant of time.
    So in your problem final displacement would be 20 ie x2(t) and t would be .... ?
     
  6. Jan 13, 2007 #5
    There's several ways you can solve this; but a good way to solve this would be to find the time it takes for it's velocity to get to 0 m/s (once it reaches it's maximum height). *Note* The amount of time it takes to go up, is __ (lower, high, or the same? You think about it :)) the amount of time it takes to go down at a given point. Then from there, to find out the total time from that given point to go 20 meters and hit the ground.

    Then from there, solve using time & distance and you can certainly get it's velocity the object needs to be thrown at =).
     
  7. Jan 13, 2007 #6

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    As said above, there are various ways to approach this question. However, one jumps out to me as being the simplest!

    Consider the total motion of the ball. How many of the variables do we know? Well, xi= 20, xf=0 u=initial velocity(to be found), a= -10, t=3. Do you know an equation relating these quantities?
     
  8. Jan 13, 2007 #7
    So, I solved for the position equation with the variables that cristo gave me and I get an initial velocity of 25/3 m/s. What amazes me is how this is SO much easier than the method that my professor used. Somehow he decided it was necessary to break the fall of the first ball into three parts and he also used the equation velocity squared is equal to twice the acceleration times the distance. Anyways, thanks again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Constant Acceleration Problem
Loading...