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Impact velocity from given height

Determining how fast something will be traveling upon impact when it is released from a given height. Created by Sal Khan.

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• If the fall was infinite, could we reach an infinite velocity ?
• The answer is no whether you fall through air or space (as in vacuum). In air, the falling object would reach terminal speed. In space, the falling object would be limited to under the speed of light.
• Sal said we were trying to find the velocity before it hit the ground. Wouldn't the final velocity be when it actually hit the ground?
• Think of it as the velocity at the instant it touches the ground before it has actually started to come to rest.
• why not just use s=1/2at^2 solve for t than use V=at?
• I think that would be another way of doing it. Worked it out and it seems like you would get the same answer...
• I am confused about the terminology used to describe acceleration and I need some clarification. If you are using a coordinate system where upwards is positive and you throw some thing upward is acceleration positive or negative? If upwards is positive then when you throw something up the acceleration is constantly increasing which to me seems incorrect. I was under the impression that if you throw something upwards the longer it's in the air the lower the acceleration until it starts falling back done and increasing again. Could someone please clarify this for me??
• If you label the up direction as positive and the force of gravity is down then the acceleration due to gravity is negative.

When you have up as positive and you throw an object up its initial velocity is positive and once it leaves your hand the acceleration of the object is only due to gravity and is constant at -9.8 m/s^2. This vertical acceleration doesn't change.

As this object travels its initial velocity will change by -9.8 m/s every second. Let's say the object was thrown up at 29.4 m/s. So since the object was thrown up which a positive direction it is initially traveling at + 29.4 m/s.

After 1 second we know that the velocity changed by - 9.8 m/s so at this point in time the object is traveling at a velocity of (+ 29.4 m/s) + (- 9.8 m/s) = + 19.6 m/s.

After another second, a total of 2 seconds, the velocity will have changed by another - 9.8 m/s so that the velocity would be (+ 19.6 m/s) + (- 9.8 m/s) = + 9.8 m/s.

After another second, a total of 3 seconds, the velocity will have changed by another - 9.8 m/s so that the velocity would be (+ 9.8 m/s) + (- 9.8 m/s) = 0 m/s.

After another second, a total of 4 seconds, the velocity will have changed by another - 9.8 m/s so that the velocity would be ( 0 m/s) + (- 9.8 m/s) = - 9.8 m/s.

And so on until the object lands.
• Since the motion of the object and the acceleration (gravity) is in the same direction, aren't you supposed to take + 9.8 m/s^2 ?
• So how do you find the time? Just plug in the Vf in V/a?
• For the final question I got -35.64 kilometers per hour and 22.15 miles per hour.
• How is 9.8 m/sec^2 the affect gravity has on an object? Doesn't it start slower than that and start speeding up until it's faster than that?
• The velocity starts at 0 m/s, and then continues to increase as long as the object is falling. So, yes, the object does start slower and will eventually reach speeds faster than 9.8 m/s, but that is velocity and not acceleration. The acceleration remains constant, and is reflecting the increasing velocity of the object as it continues to fall. The object won't stop accelerating until it hits the ground.
• Why is the acceleration -9.8 m/s^2. Shouldn't it be a positive since you are throwing the rock down?
• The convention in the video is that: down is negative.