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Terminal Velocity Question

  1. Jan 4, 2006 #1
    I know terminal velocity is when the upward force is equal to the gravitational pull, but is that only in free fall? Would something with air resistance like an arrow shot from a bow be affected by terminal velocity?

    I'm not to clear on how it works...
  2. jcsd
  3. Jan 4, 2006 #2


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    Welcome, Applesauce. An arrow is in free-fall as soon as it leaves the bow. Unless shot vertically, though, it won't have time to reach terminal velocity on the way down. Everything that isn't under power of some sort has it's own unique TV, which varies with aerodynamic qualities and air density. That's about all I have to contribute, but others can give you more information.
  4. Jan 4, 2006 #3
    Right, which means that the force of gravity is acting on it, so the forces can't sum to be zero on it's way up.

    You see, terminal velocity happens when the drag force from collisions with air molecules (an upward force) cancels out the downward force of gravity. So, if the arrow was shot upward really fast then there would just be extra deceleration on it because the drag force would be down.
  5. Jan 4, 2006 #4
    Let say something is travelling horizantaly in the air in constant acceleration. The higer the velocity, the higher the viscous drag is.Thus the acceration decreases throughout time. Eventually the acceleration becomes zero and the object travels with constant velocity. The velocity is called terminal velocity.
  6. Jan 4, 2006 #5
    But the arrow wouldnt have a constant velocity right?

    I'm still really confused on terminal velocity. Any one have a real world example?
  7. Jan 4, 2006 #6


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    Correct, but that's just because it isn't at terminal velocity yet.
    Let's stick with the arrow, but to make terminal velocity happen, lets fire it off a cliff, in a high arc. The arrow is essentially in freefall even when going up - the only forces acting on it are drag and its weight. The drag force works to decrease both the upward and forward velocity while the weight only decreases the upward velocity. Eventually it pitches over at the top of its arc and starts down. Now the drag force is working to slow its downward acceleration while still slowing its forward velocity. Its weight is causing downward acceleration. Eventually, the horizontal component of the drag force stops the forward speed and the arrow is only moving downward. And eventually again, it accelerates to a velocity where the drag force comes into equilibrium with its weight and stops the acceleration - terminal velocity.
    Last edited: Jan 4, 2006
  8. Jan 4, 2006 #7
    Hmmm I understand. But I feel as if I may have missed something.

    The drag will eventually "cancel out" the acceleration due to gravity after falling for some time... I understood that. But not too sure as to why though.

    Does it have to do with Newton's "equal and opposite reaction" theory? That the free-falling arrow's velocity will become so fast, that the opposte reaction (Being the drag/atomsphere) pushes back up with equal force, effectively cancelling out any further acceleration?

    That means the drag will eventually have a constant acceleration of 9.8m/s/s then...
  9. Jan 4, 2006 #8
    By the way that avator is really scary, Danger.

    Is it a monkey?
  10. Jan 5, 2006 #9

    D H

    Staff: Mentor

    Drag acceleration increases with the square of the velocity. Drag thus increases dramatically as a falling object builds up speed. Imagine a plot of drag acceleration versus velocity. The terminal velocity is simply the velocity at which the drag acceleration is equal to the gravitational acceleration.

    To demonstrate this consider an arrow released from a stunt airplane.

    1. The plane climbs vertically. The arrow is released at the top of the climb such that the arrow's initial velocity with respect to the ground is zero. The arrow will fall toward the Earth, accelerating until it reaches a terminal velocity.

    2. The plane goes into a forced dive. The arrow is released during the dive such that the arrow's intial velocity is faster than the terminal velocity. The arrow will again fall toward the Earth, this time decelerating until it reaches a terminal velocity.

    Terminal velocity has nothing to do with Newton's "equal and opposite reaction" theory. Consider an airless planetoid such as our moon. An arrow (or golf ball) flies a perfect parabolic arc when shot (or driven) from the surface of the moon.
  11. Jan 5, 2006 #10


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    Drag is a force, not an acceleration. But you can use net force to calculate acceleration: By adding the forces to find the net force on the arrow (0) and applying f=ma, you can find the acceleration (0).
  12. Jan 5, 2006 #11


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    It's me, on Hallowe'en last year. :biggrin:
  13. Jan 10, 2006 #12


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    The initial velocity may be higher than terminal velocity. Assuming that the arrrow doesn't hit anything, eventually it starts heading straight down, and it's speed decreases or increases until the aerodynamic drag matches the gravitational pull.

    For a human skydiver with a typical jump suit in a semi-prone position (leg's bent a little), terminal velocity is about 120mph.

    Note that there can be other forces as well, even a propelled object will reach terminal velocity when aero-dynamic drag forces equals the propelling force which can include gravity.

    Any form of energy will work. For example, two air streams flowing at different speeds realitvely close to each other can be used as a power source for radio control gliders. Typically this happens on the downwind side of a ridge, where there's a near horizontal flow of air just above a large pocket of air that is barely moving or possibly moving the opposite way due to circulation. Taking advantage of this is called dynamic soaring. The most recent record for this is 301mph (as captured by a radar gun), well above the free fall terminal velocity for a glider.

    If interested, a video was made of the 301mph run (although zoomed in too close), plus some other dynamic soaring videos where you can get a better look, at this web page:

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