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Conservation of energy

  1. Aug 15, 2005 #1
    If a 3 kg ball is thrown straight up at 40 m/s, using energy conservation, calculate how high the ball would go if there was no wind resistance.

    I was told this could not be calculated. Is that true? and if so, why?
  2. jcsd
  3. Aug 15, 2005 #2


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    why not, loss in KE = gain in PE ?
  4. Aug 15, 2005 #3
    What is the formula?
    What if I added, lets say, a max. height like 75 m. Could energy lost, like due to friction with the air on the way up be calculated?
  5. Aug 15, 2005 #4
    This can be done by looking at a symmetric problem. Assume the ball does decelerate to 0. Then by symmetry, solve the problem of how far it would have had to accelerate under gravity near the Earth's surface to get to 40 m/s, since assuming it loses no energy to air resistance this is the velocity it would return with to the person's hand.
    If you want to add in air resistance, you have to account for the cross-sectional area of the ball (A), as a sheet of paper will fall slower than a crumpled up piece. Similarly, a styrofoam ball will encounter more resistance than a rubber ball, so density (d) is inversely proportional to the air resistance. Next, stick your hand out when you're walking. Compare it to sticking your hand out when you're running or in a car. Speed (v) is thus proportional to air resistance. We're insterested in the distance the ball will travel. m is the mass of the ball. So far we have the rudimentary 1-dimensional net force equation ma = mg - A*d*v. But v is not a constant, it is a function of time, so we have to solve the differential equation m*dv/dt = - mg - A*d*v for speed v or m*d2x/dt2 = - mg - A*d*dx/dt for distance x. This simple model has many more refinements placed on it based on the speed of the object and the type of resistance, but solving it will already allow you to see simple behavior like terminal velocity. :smile:
    Since the energy lost by the ball is equivalent to the work done by the air resistance on the ball, you want to calculate the total work done by air resistance in the above equations, which would require integration, since the air resistance is also a function of time.
    Last edited: Aug 15, 2005
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