If an object is moving at terminal velocity and then opens the parachute, the air resistance is higher than the weight so why doesnt he go upwards.

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    $\begingroup$ This misunderstanding is tightly coupled with that in your previous question. They are both resolved by understanding that force is related to acceleration and not to velocity. So, if you are not clear on the exact relationship between acceleration and velocity it would be worth your time to focus on that difference for a while and then re-visit these force questions after you have mastered the kinematics issue. $\endgroup$ – dmckee --- ex-moderator kitten May 6 '18 at 17:48
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    $\begingroup$ You may have seen film or video in which a free-falling sky diver pulls the rip cord and suddenly seems to shoot up into the sky. But that's only because the one who pulled the rip cord suddenly was falling much more slowly than the skydiver who was holding the camera. From the camera's point of view, the one who pulled the rip cord seems to be going up, but from the point of view of a person standing on the ground, they're both still falling down. $\endgroup$ – Solomon Slow May 6 '18 at 18:06

Because force and motion aren't always in the same direction. Suppose you slam on the brakes in your car. Your car doesn't instantly go backwards. Instead, it accelerates backwards.

When a skydiver opens their parachute, they accelerate upwards. This is not the same as saying the move upwards.

Acceleration is a change in the velocity. So the amount of downward velocity the skydiver has gets smaller and smaller, until they reach a new terminal velocity. During this time, the change in their velocity is upwards.

If we illustrate the skydiver's velocity in red and force and acceleration in blue, we might get something like this:

enter image description here

Time goes from left to right. The skydiver opens their parachute and feels and upward force. It pushes upward on them, but isn't enough to completely stop them and make them move upward. Instead, it just slows them down a bit as they continue falling downward. They keep feeling an upward force (while moving downward) that slows them until they get to a new terminal velocity.

Note that this picture shows the velocity arrows next to each other in space; in reality, the skydiver gets lower and lower in space as we move to the right, but that was harder to draw.

  • $\begingroup$ In your diagram, why doesn't the acceleration decrease to zero as the velocity decreases? Instead, it seems to increase. $\endgroup$ – sammy gerbil May 7 '18 at 17:07
  • $\begingroup$ That's because it's just a rough sketch; it's not intended to show as much detail as you're asking about. $\endgroup$ – Mark Eichenlaub May 8 '18 at 1:01

Force can oppose motion. Parachutes provide upward acceleration but the acceleration downwards is much larger in the bigger picture. This is not the same as saying the move upwards; parachutists acceleration > acceleration due to parachutes.

Acceleration is a change in the velocity**

downward velocity eventually decreases to match air resistance and they reach a terminal velocity phase where both upward and downward forces are equal and the parachutist will fall at the same rate.

unless there's an upward force that's much larger than the downward force on a stationary object, the object will not move upwards.



Something going forward will have to slow down before going backward. For example a boat driving forward at 15 km/h has to slow down to 0 km/h before it starts going backward. Same thing for the man. However as he slows down the upward force, in this case drag, produced by the chute falls, meaning it will eventually drop to a level where drag = gravity, and the man will fall at a constant rate. Therefore the man will never begin rising, unless there is some external force like an updraft or a gust of wind.

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Summing it up, It's my conclusion that: 1) If an object is at rest, a car for instance; and a force is applied it will accelerate until the force is applied. Now we stop exerting the force and it will continue its motion with net force but since in our world friction force is there; it's impossible. 2) Secondly, parachutists don't move upwards or backward because the force is not that large enough. As said earlier by referring to a car, first it decelerates and if the Force is big enough it moves in the opposite direction. That's why the parachutist's speed keeps on decreasing until Weight=Air Resistance. I came up with this conclusion. Anyone please check it and rectify if needed. I want to make sure, it's correct. Thanks

  • $\begingroup$ Hey Abdullah, welcome to Physics Stackexchange! Please try to re-order your ideas and rephrase them, it is very hard to follow your thoughts... $\endgroup$ – 2b-t Mar 18 '20 at 0:19
  • $\begingroup$ In case you are interested if your way of thinking is correct, here is a short video about parachuting and the corresponding forces: youtube.com/watch?v=BpXlz4XWpKA Note that the forces acting upon the body are only connected by Newton's second law to the acceleration and not the velocity. $\endgroup$ – 2b-t Mar 18 '20 at 0:26
  • $\begingroup$ @2b-t Thanks, I really like this forum. Secondly, I've got the answer to my second point for the parachutists.... But about my first point,could you please clarify: $\endgroup$ – Abdullah Khalid Mar 18 '20 at 9:31
  • $\begingroup$ @2b-t Thanks, I really like this forum. Secondly, I've got the answer to my second point for the parachutists. But about my first point, could you please clarify: All I'm saying is that, If an object for e.g: a car is at rest and we exert a force on it; it will accelerate until we keep applying that force,10N let's say. But when we stop applying10 N Force, it should continue its motion with zero acceleration such as in Vacuum but since friction Force is there on earth, the car will come to a halt. I hope I was able to make an explicit statement this time! $\endgroup$ – Abdullah Khalid Mar 18 '20 at 9:42
  • $\begingroup$ You can learn a lot here by reading and contributing, there are a lot of smart people around here. The way you put it now is correct but your initial comment is very hard to understand. Try to rephrase it, add formulas (such as Newton's second law) and format them with MathJax (a special version of Tex very similar to LaTeX) math.meta.stackexchange.com/questions/5020/… and add links by typing [link description...] (mylink.com) without the space in between. Sometimes a bit of well written math is easier understood than a long text... $\endgroup$ – 2b-t Mar 18 '20 at 11:14

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