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Trajectories of objects thrown from a height

  1. Aug 3, 2012 #1

    I am new to this forum. I am having trouble with conceptualizing trajectories of objects.

    Here is a specific question that I can figure out:

    If a car is travelling down the hill with an incline and an object is thrown out of the car, what is the trajectory of the object? Would the object move in a half inverted parabola shape? How can I tackle such questions?

    Another question on kinetic energy and momentum conservation:

    A 2.00 kg mass is moving at 2.00 m/s and strikes another 2.00 kg mass which is stationary. If kinetic energy is conserved, how will they move following the collision. Assume that the fist mass is moving to the right.

    My guess is that the second will move to the right with a speed of 2m/s while the first mass will become stationary. I am not sure and would appreciate if someone could help explain this.

    Last edited: Aug 3, 2012
  2. jcsd
  3. Aug 3, 2012 #2
    Welcome to PhysicsForums ostrich!

    Trajectories comprise a lot of different stuff. Firstly, you are correct in assuming that the object will move in an inverted parabola. A good source for general knowledge (including the general equation of motion) is here: http://en.wikipedia.org/wiki/Trajectory

    If you are looking for simple values, this is a good website that can calculate various values you might be looking for: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

    In general, you can approach the problem in a couple different ways. If you know calculus, you can start by knowing that the acceleration due to gravity is a constant no matter what you are talking about. From there you can anti-derive to figure out the equations of motion.

    If you don't know calculus, I would start with the equations listed on the pages I linked. In general, all you really need to figure out the trajectory of any object is the angle ##\theta## and the velocity ##v## of the object in question, and using the various equations you can figure everything else out.

    Let us know if you need any more help :)
  4. Aug 4, 2012 #3
    Any motion under gravity is a sum of two motions: the motion in a straight line with a constant speed (determined by the velocity of the object as it left the car - direction is important!) and the vertical accelerated downward motion due to gravity. For motions within relatively small regions, we can assume gravity to be constant, then the downward motion is uniformly accelerated, in which case the resultant motion is a parabola.

    Regarding the other question, your intuition is right. This could be proven formally by applying the laws of conservation of momentum and energy.
  5. Aug 4, 2012 #4


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    Presumably what you are trying to find out is the trajectory of the ball relative to some reference point (eg an observer standing on horizontal ground).

    What happens to the car before and after the ball is released doesn't matter. "All" you need to do is work out the initial velocity and direction of the ball with respect to the reference point on horizontal ground.

    Lets say the problem gives you information about the initial path of the ball with respect to the car. You have use that, and information on the path of the car, to calculate the initial path with respect to horizontal ground. Typically that would involve vector addition.

    Once you have calculated the initial conditions for the ball the problem becomes similar to that involving a ball fired from a stationary cannon.
  6. Aug 4, 2012 #5
    Thanks for the quick response!

    What I was getting confused with is that this question stated to draw out the trajectory of the object but also mentioned that the object is thrown out of the the can while the car is coming down on the hill. I know that stationary objects thrown from a height follow an inverted parabola shape similar to projectile motion when objects are thrown from a building but what if the car is moving while the object is thrown out of it, does that change anything in this situation?
  7. Aug 4, 2012 #6
    Resolve the velocity of the down hill to horizontal and vertical.
    These velocities will be the initial vertical and horizontal velocity of the object.
    The equation still an inverted parabola.
  8. Aug 4, 2012 #7


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    Yes it does change things. For example if you are driving in a car at 30mph and you throw a baseball backwards at 30mph (with respect to the car) then it's forward motion is zero relative to the ground and it's just going to drop vertically.
  9. Aug 4, 2012 #8
    It changes things, but a ball will still always fall in a parabola (if you let me call a straight line an infinitely skinny parabola). The difference comes with the initial velocity.

    If you know how to add vectors, adding the vector for the velocity you threw the ball with and the velocity of the car, the resultant vector will be the proper velocity to enter into the equations.
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