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I recently read about the property of terminal velocity for objects and I got a question when doing so.

If from a very tall building I throw a ball faster than terminal velocity downwards, will the ball slow down, then continue with terminal velocity, or will it continue with the speed I threw it with?

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The ball will slow down to terminal velocity. This is because the force of air drag increases with increasing speed. Terminal velocity is the speed where the force of air drag equals the force of gravity, so the total force is zero and the object travels at a constant speed. If the ball has a higher speed, then it will have an air drag force greater than gravity and slow down.

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    $\begingroup$ Note that you don't need a tall building or super strong arm to test this. You can test it at home with any object that has a sufficiently low terminal velocity (feather, object attached to a parachute, etc.) $\endgroup$ – DreamConspiracy Mar 6 at 17:04
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    $\begingroup$ @DreamConspiracy A balloon works great for this $\endgroup$ – slebetman Mar 6 at 19:24
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    $\begingroup$ Important caveat: this assumes that the ball has enough time to slow down to terminal velocity. A sufficiently-dense ball, or one thrown with sufficient speed, may impact the surface while still travelling appreciably faster than terminal velocity. $\endgroup$ – Vikki - formerly Sean Mar 6 at 23:56
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    $\begingroup$ @Vikki Good point, but it will still impact slower than it was thrown. $\endgroup$ – reirab Mar 7 at 8:57
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    $\begingroup$ It's worth keeping in mind that terminal velocity isn't just a function of mass and gravity. It's also surface area. Which is why you shouldn't forget to deploy your parachute and change your terminal velocity. $\endgroup$ – candied_orange Mar 7 at 13:10
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Terminal velocity is achieved when the drag force $f=Dv^2$ (where $D$ is a constant and $v$ is the object's speed) equals the gravitational force $F_g=mg$. Equating these forces is the condition for terminal speed, which is $\displaystyle v_T=\sqrt{\dfrac{mg}{D}}$.

If the ball is thrown with an initial velocity greater than the terminal velocity, then $f>F_g$, so the ball will slow down until $f$ decreases enough such that $f=F_g$, in which case the forces are balanced and terminal velocity is achieved.

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    $\begingroup$ It might be illuminating to derive an equation for terminal velocity in terms of an initial velocity. Traditional equations for terminal velocity are derived assuming the objects falls from rest. However, we could incorporate an initial velocity, find the instantaneous velocity as a function of time and initial velocity, and take the limit to obtain terminal velocity in terms of initial velocity. It might be interesting to inspect how the terminal velocity changes mathematically when the initial velocity is made large. $\endgroup$ – electronpusher Mar 8 at 5:32
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A similar situation occurs when you push hard on a door in between 2 rooms (all other doors and windows are shut) and it slows down until it reaches the terminal velocity.

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    $\begingroup$ Slow down then, will not, theoretically reach terminal velocity in very large time (theoretically infinite time) . So it depends on the length of the building, how percentage the difference between earth collision speed and the terminal speed relative to the terminal speed. $\endgroup$ – Ahmed Kamal Kassem Mar 6 at 10:09
  • $\begingroup$ @AhmedKamalKassem I know what you mean but I want to resist the assumption that exponential decay implies infinite time. Rather, it takes a finite time to reach a velocity where the remaining difference from terminal velocity is small compared to influences that are neglected in a simple treatment that predicts exponential decay. After that finite time, the velocity will fluctuate as the ball is buffeted by currents of air etc. $\endgroup$ – Andrew Steane Mar 6 at 11:00
  • $\begingroup$ @Andrew Steane I understand, and I see your claim is more practical than mine $\endgroup$ – Ahmed Kamal Kassem Mar 6 at 15:57
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    $\begingroup$ This sounds more like air pressure equalisation? $\endgroup$ – Criggie Mar 7 at 0:17
  • $\begingroup$ There is no terminal velocity of a closing (vertical) door. This does demonstrate drag (with some additional effects from air pressure in the rooms as well as friction in the hinges,) but not drag slowing a falling object, since a door is generally not a falling object. Also, this seems more like a comment than an answer to the question that was asked. $\endgroup$ – reirab Mar 7 at 9:00
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An object with initial velocity greater than it's terminal velocity will decelerate until it finally hits the ground. Terminal Velocity, NASA

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