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Throwing objects in the air and calculating height problem

  1. Aug 23, 2012 #1
    Hey folks I'm having some trouble with this question in my homework:

    1. The problem statement, all variables and given/known data

    A man at the top of a building throws an object straight up at a speed of 10 m/s. The object hits the ground 5 seconds latter. For the purpose of this problem the assumed environment is a vaccum.

    a) What is the highest point the object reached?
    b) What's the height of the building?
    c) At what speed does the object hit the ground?



    2. The problem

    How does the 10 m/s speed relate to the 9.8 m/s^2 gravitational acceleration, does it just slow down the object at -9.8 m/s^2 ?
     
  2. jcsd
  3. Aug 23, 2012 #2
    I may be wrong here but if the initial velocity will as it states the question cause the object to oppose the force of gravity. As the object moves up due to the force that the man threw it up, gravity is acting on it. Eventually gravity will slow the object down to 0 (at its heighest point above the ground where its kinetic energy will be 0 and its potential energy will be its greatest) and start bringing it back to Earth until the object eventually hits the ground and is stopped by the ground.

    Just trying to help, hopefully someone can verify my idea. You may wish to take a look at: http://www.physicslearningsite.com/projectile.html
     
  4. Aug 23, 2012 #3
    Hello ,
    As the man imparts kinetic energy to the object , The body starts moving up . But as gravity is acting downwards , its KE will start to decrease till it becomes zero , which would be at the highest point . Now at the highest point the particle will do a free fall and reach the ground . During the free fall the KE will increase , and Potential energy due to gravity will decrease .
    If you take the regular sign convention , we can take the initial velocity to be positive and gravity as negative , which means the same thing as gravity is in opposite direction to velocity .
     
  5. Aug 23, 2012 #4

    CAF123

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    Gold Member

    To start you off, the object will reach it's highest point when [itex] v_y = 0.[/itex]Using this condition and another kinematic relation, you should be able to find the highest point. (above the point of release)
     
  6. Aug 23, 2012 #5
    For (b)

    Total time = 5secs.
    Initial velocity up.
    For conservation of energy, the magnitue of velocity up is equal to magnitude of velocity down at point s=0

    Find the time taken for the object to return to same level as it was thrown, that is s=0.
    Then you have time to reach the ground and its inital velocity
     
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