There is no case where the ball could be traveling faster on the way down if there are damping forces present. The fastest possible downward speed it can have is $v_T=316.2$ m/s, and in the case of an initial speed of $v_0=100$ m/s, it will have reduced its speed to approximately $70$ m/s. It's actually an optimistic estimate, considering that in the presence of air drag, the ball will not actually make it to the same height it would in the absence of air drag. In fact, you will notice that the amount of time spent in the air is actually less than the "ideal" (no air drag) time in the air of $10$ s. Which gives a time to fall from max height of approximately $t=7$ s. Which is a nonlinear first order differential equation. If we consider the downward motion of the projectile, we can use Newton's 2nd Law to obtain the equation: ![]() So if you throw it upward at $v_0=100$ m/s at a height of $0$ m, without air drag it will be moving at a speed of $100$ m/s when it falls back into your hand. Beyond a certain launch speed, it always lands at the same speed.Īt best, without air drag, your projectile will be traveling at the same speed at any given height, regardless of the direction. Flooding in the plan to withdraw from earth. Lets keep a little faith, always be by the side flying far away, hands holding children far away. Grown up, the fragrant smile is still on the lips. So not only is the final speed at the bottom always lower than the launch speed, it tends towards a maximum value. Gently let the childhood fly innocently, who passes the time, we fly forever to the far away skies, open the door and see what the world is afraid of. Note also, that if it is launched high enough, the air resistance on the way down (which increases as its speed increases) will eventually balance the acceleration due to gravity and it will reach terminal velocity. So it is moving slower than when it left. It finally arrives back where it started with $E_f < E_p < E_0$. It starts back down again, converting $E_p$ back into kinetic energy, but also losing some more energy as it pushes the air away again. At the top, the ball stops and now has $E_p < E_0$. air resistance - it is converted to kinetic energy in the air molecules it pushes away). it is converted into potential energy) and some of it is spent on moving the air out of the way (i.e. Some of this energy is "spent" in lifting the ball (i.e. The ball starts off with a kinetic energy $E_0$. This is simply because it must lose energy to air resistance during its flight. ![]() The ball will always return to the ground at a lower speed than it was launched.
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