Say a baseball is shot from a batting machine in a straight line (I don't know much about baseball, but hopefully this makes sense). Normally, when demonstrations on kinetic and potential energy are done, it uses a vertical movement. An example is a ball being thrown in the air in an arc and being pulled back down by the force in gravity. However, if a ball, or simply projectile, is shot in a straight line, when would the kinetic energy be the greatest and when would the potential energy be the greatest? Of course, this also factors in the force of gravity pulling down on the projectile. Thanks!
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$\begingroup$ If by "in a straight line" you mean horizontally, then the potential energy will be greatest immediately after leaving the machine, and the kinetic energy will be greatest immediately before hitting the ground. $\endgroup$– Riley Scott JacobCommented May 15 at 0:08
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$\begingroup$ When you say measure, do you mean actually measure it or calculate it? These are two completely different ways to approach the problem. $\endgroup$– TriatticusCommented May 15 at 0:26
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1$\begingroup$ A projectile shot out of a batting machine does not travel in a straight line. It may start out horizontal, but gravity immediately starts pulling it downward. The downward velocity gets faster and faster. It travels in a curve. $\endgroup$– mmesser314Commented May 15 at 0:48
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1 Answer
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The potential energy of the baseball is given by: $$PE = mgh.$$ Note that $m$ and $g$ are both constants. Therefore, the baseball’s potential energy will depend only on the height of the ball above the ground. The moment it is launched by the batting machine (presumably horizontally) gravity will begin to accelerate it toward the ground, causing it to lose some altitude and therefore lose some potential energy. Therefore, the moment immediately after the ball is launched is when it will be at its maximum height and therefore have its maximum potential energy. The acceleration will also cause the ball to gain some velocity in the downward direction. Assuming negligible air resistance, the horizontal velocity that the ball began with when it was launched doesn’t change over the course of its trajectory, because the only force acting on the ball is that of gravity which has no horizontal component. So the ball hits the ground with its initial horizontal speed and some additional vertical speed from gravity (which increases all the time the ball is in the air). Since kinetic energy is given by: $$KE = \frac{1}{2}mv^2,$$ the time when the ball has the greatest kinetic energy is the time when it has the greatest total speed. Since the total speed only increases as the ball falls, the moment of highest kinetic energy is right before the ball hits the ground. Something to note: if gravity is the only force acting here (i.e. no friction), then the amount of potential energy lost as the ball fell is equal to the amount of kinetic energy it gained from the acceleration due to gravity: $$\Delta{PE}=-\Delta{KE}.$$
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$\begingroup$ Thanks! This was very helpful and your answer makes a lot of sense. Just to be clear, however, we still have the same amount of total energy in the system, correct? There is simply a conversion between potential and kinetic energy? $\endgroup$ Commented May 15 at 14:39
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$\begingroup$ That's exactly correct. The last equation in my answer states that all the potential energy that was lost is equivalent to the kinetic energy that was gained. So the net change in energy of the baseball is zero during its trajectory. $\endgroup$ Commented May 16 at 17:16