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!

Free Fall Ball Question

  1. May 13, 2007 #1
    A ball is thrown upward from the top of a 25.4-m-tall building. The ball's initial speed is 12 m/s. At the same instant, a person is running on the ground at a distance of 32.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

    This problem will use the kinematics equations.

    Here is what I have so far:

    First, I will want to determine the time that the ball will spend in the air. Then, I can divide the distance by the time and that will give me the average speed of the person.

    First, I must account for the time the ball will spend going up after being thrown. I calculated this to be:

    V = Vinitial + at
    0 = 12.0 + (-9.80xt)
    t- 1.22449

    I can then multiply this number by 2 to account for the ball coming back down to the "top of the building" So far, total time is 2.44898.

    Now I must account for the time the ball is falling down the 25.4 m.

    Using this formula: displacement = vinitialt+ .5at^2
    -25.4 = 12.0t + .5(-9.8)t^2
    0 = .4.9t2 +12.0t + 25.4
    I solved for this, and found the time to be 3.80965 seconds.

    That would give me a total time of 3.80965 + 2.44898 = 6.25863

    Then, Avg Speed = Distance/Time = 32.0/6.25863 = 5.11924 m/s

    This answer is not correct though, and I am having trouble figuring out what I am missing.

    Thanks for the help
     
  2. jcsd
  3. May 13, 2007 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This bit is correct. There is no need to add on the other times you calculated, this calculation takes both the upward and downward motion into account.
     
  4. May 13, 2007 #3
    Oh, I see now. Thanks so much for the help.

    If you do not mind, I have another two questions that I would really appreciate someone answering if they had the time.

    A hot-air balloon has just lifted off and is rising at the constant rate of 2.4 m/s. Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward. If the passenger is 1.4 m above her friend when the camera is tossed, what is the minimum initial speed of the camera if it is to just reach the passenger?


    So, the first thing I thought I should do was calculate the distance the ballon and camera would be traveling.

    The balloon would have been:
    d = vinitialt + .5at^2
    d = 2.4t

    Now, the distance traveled by the camera would be:
    d + 1.4 = 0(t) + .5(9.8)t^2
    So, if i substitute for d, I can get thsi equation
    2.4t + 1.4 = 0(t) + 4.9t^2
    0 = 4.9t^2-2.4t-1.4

    Solving for t gave me the value of 0.83285.

    Now, to find the initial speed, I know that
    a= (vfinal-vinitial)/t

    Therefore,
    -9.8 = 0-vinital/.83285

    v initial would then equal 8.16193 m/s but this isn't correct. I am thinking that I might have mixed up a sign somewhere maybe?

    Also, this I am sure is a simple question, but this is my first time with physics and I just cannot get anything going that is correct.

    So, if I am traveling at 30m/s and want to pass a truck, in order to do that I need to cover 36m, and my acceleration is 2.7 m/s^2, what formula can I use.

    I had thought this:

    36 = 30(t) + .5(2.7)t^2
    36 = 30t + 1.35t^2
    1.35t^2 + 30t - 36 = 0
    t=1.1414 which is not.

    I am getting hung up on the fact that I am not acclerating from zero. I guess I need to now how to account for the fact that I am already traveling at a speed of 30m/s.

    Thanks again
     
  5. May 13, 2007 #4

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You are taking it into account. That's what the term in red is for. Can you give us the exact problem statement?
     
    Last edited: May 13, 2007
  6. May 13, 2007 #5

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you are trying to figure out the initial speed of the camera, why have you set it to zero here???

    Also, the acceleration due to gravity is downward. You have it positive here, but you're using a coordinate system with a positive upward convention.
     
  7. May 13, 2007 #6
    Cepheid,

    Thank you for your responses. The exact question for your first reply was:

    A car is behind a truck going 30.0 m/s on the highway. The driver looks for an opportunity to pass, guessing that his car can accelerate at 2.7 m/s2, and he gauges that he has to cover the 16.0 m length of the truck, plus 10 m clear room at the rear of the truck and 10 m more at the front of it. In the oncoming lane, he sees a car approaching, probably traveling at the speed limit of 25 m/s. He estimates that the car is about 500.0 m away.
    How long does it take the driver to complete the pass?

    If x1 is the position of the driver, and x2 is the position of the oncoming car, what would be the value of separation distance (x2 - x1) at the end of a successful pass?
    After I know how long it will take, I will be able to calulate the distance the second car would have traveled. Then, from that number, I was planning on substituing the 36m the original truck would have gone.

    c) What would be the driver's speed at the end of the pass?


    With regards to your second reply, the d + 1.4 = 0(t) + .5(9.8)t^2 was set at zero as I saw that being the time when the two distances the camera and balloon would have traveled would have been equal and therefore the zero represents the final velocity of the camera ( as it just barely made it into the person's hands) From this, I was planning on backtracking and then obtaining the initial velocity after I knew the time until the distances were equal.

    Thanks.
     
  8. May 13, 2007 #7
    So, when i posted my answer to the following equation
    A hot-air balloon has just lifted off and is rising at the constant rate of 2.4 m/s. Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward. If the passenger is 1.4 m above her friend when the camera is tossed, what is the minimum initial speed of the camera if it is to just reach the passenger?


    So, the first thing I thought I should do was calculate the distance the ballon and camera would be traveling.

    The balloon would have been:
    d = vinitialt + .5at^2
    d = 2.4t

    Now, the distance traveled by the camera would be:
    d + 1.4 = 0(t) + .5(9.8)t^2
    So, if i substitute for d, I can get thsi equation
    2.4t + 1.4 = 0(t) + 4.9t^2
    0 = 4.9t^2-2.4t-1.4

    Solving for t gave me the value of 0.83285.

    Now, to find the initial speed, I know that
    a= (vfinal-vinitial)/t

    Therefore,
    -9.8 = 0-vinital/.83285

    v initial would then equal 8.16193 m/s but this isn't correct. I am thinking that I might have mixed up a sign somewhere maybe?

    the 8.16193 answer I was told was off by orders of magnitude. I understand an order of magnitue of 1 is like from 1 going to 10 but do not understand how it applys to this problem
     
  9. May 13, 2007 #8

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Again, you still haven't answered my question. Why have you set the initial velocity to zero (highlighted in RED), when this is what you are trying to solve for? In other words, if d(t) is the distance of the camera above the person as a function of time, then

    [tex] d(t) = v_0t + \frac{1}{2}at^2 [/tex]

    where [itex] v_0 [/itex] is the initial velocity, the quantity you are trying to solve for. You have set it to zero, which makes no sense!

    Also, (another thing I pointed out before), the acceleration is negative.
     
    Last edited: May 13, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Free Fall Ball Question
  1. Ball in free fall (Replies: 4)

  2. Ball in free-fall (Replies: 3)

  3. Ball free fall (Replies: 4)

  4. Free-fall of a ball (Replies: 2)

Loading...