1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Opposing accelerations on stone drop/throw up problem

  1. Mar 27, 2013 #1
    1. The problem statement, all variables and given/known data
    A stone is dropped from the top of a 50m high tower at the same moment as a ball is thrown straight upward from the bottom of the tower, with an initial velocity of 20m/s. when and where do the ball and stone pass each other?

    2. Relevant equations
    I actually have the solution in front of me. My frustration comes when it identies the given variables. The acceleration of the thrown ball is negative as the velocity is positive. This means that down is negative. In the solution, the dropped stone is assigned a positive acceleration. To me the stone is speeding up in the negative direction, isn't that negative acceleration? How can each object have separate direction's considering they are both used in the solution?

    3. The attempt at a solution
  2. jcsd
  3. Mar 27, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Hello Tallgirl. Welcome to PF !

    Can you give the two equations, exactly as they're written in that solution?
  4. Mar 27, 2013 #3
    Try following your method (stick to the sign convention) and see whether you get a different answer.

    (Note: If you take the downward direction as negative then all vectors pointing downwards, including displacement will be negative)
  5. Mar 27, 2013 #4
    Did you get the equations?
  6. Mar 31, 2013 #5
    Here's how it is given:

    Stone's variables:
    vi = 0m/s
    a = 9.8m/s*2
    t = ?
    d = .5at*2 = (4.9m/s*2)(t*2)

    Ball's variables:
    vi = +20m/s
    a = -9.8m/s*2
    t = ?
    d = vit + .5at*2 = (+20m/s)t + (-4.9m/s*2)(t*2)

    d1 + d2 = 50m

    (4.9m/s*2)(t*2) + (+20m/s)t + (-4.9m/s*2)(t*2) = 50m

    (+20m/s)t = 50m

    t =2.5s

    And then plug this into the d equations to find the distances.
  7. Mar 31, 2013 #6
    This is correct.

    But according to your sign convention, a = -9.8 m/s2, right? This would then mean, the d that you have found (0.5gt2) will also be negative. But then, d1+d2 wouldn't be the same right?

    This seemingly paradoxical statement is a result of not taking proper signs into consideration.

    Now if total distance is D (from the bottom) and d is the displacement of the stone from the high point, d would be negative. Then the displacement of the stone from the bottom would be D+d( vector sum;that means, you are taking them along with the signs). Its magnitude would be D-d and hence when the stone and ball meet, (with d2 being the ball's displacement from the bottom) D= d2 - d (vector difference). Thus since d is negative, magnitude of d2 - d is d2 + d which is D.
  8. Mar 31, 2013 #7
    Thank you so much.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted