Current time:0:00Total duration:11:43
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Determining how fast something will be traveling upon impact when it is released from a given height. Created by Sal Khan.
Video transcript
WHAT I WANT TO DO IN THIS VIDEO IS ENTER AN AGE OLD QUESTION, AT LEAST AN INTERESTING QUESTION TO ME. AND THE QUESTION IS LETS SAY I HAVE A LEDGE HERE. A LEDGE, OR CLIFF, OR BUILDING OF SOME KIND AND LETS SAY IT HAS A HEIGHT h, RIGHT OVER HERE, AND WHAT I AM CURIOUS ABOUT IS THAT IF I WERE TO EITHER, THIS IS ME OVER HERE IF I WERE TO JUMP BY MY SELF, OR I THROW SOME THING, LIKE A ROCK, FROM THIS LEDGE, HOW FAST, EITHER MY SELF OR THAT ROCK BE GOING WHEN IT HITS RIGHT BEFORE THE GROUND. AND LIKE ALL OF THE OTHER VIDEOS WE ARE DOING ON PROJECTILE MOTION RIGHT NOW, WE GOING TO IGNORE AIR RESISTANCE, AND FOR SMALL h's AND SMALL VELOCITIES, ITS ACTUALLY REASONABLE, OR THE OBJECT IS AERODYNAMIC THEN AIR RESISTANCE WILL NOT MATTER LESS, ITS ME KIND OF BELLY FLOPPING FROM HIGH ALTITUDE THEN THE AIR RESISTANCE WILL START TO MATTER A LOT. BUT FOR THE SAKE OF SIMPLICITY, WE ARE GOING TO ASSUME NO AIR, OR WE ARE NOT GOING TO TAKE INTO THE AFFECTS OF AIR RESISTANCE ASSUME WE ARE DOING THIS ON AN EARTH LIKE PLANET THAT HAS NO ATMOSPHERE. HOW EVER YOU WANT TO DO IT. SO LETS JUST THINK ABOUT A PROBLEM MY BE SOME OF YOU SAY THAT IS NOT REALISTIC BUT IT WILL BE REALISTIC FOR SMALL "h" IF YOU JUMP OFF A ONE STORY BUILDING, THE AIR RESISTANCE WILL NOT BE A MAJOR COMPONENT TO DETERMINING YOUR SPEED IF IT WAS TO BE A MUCH LARGER BUILDING, THEN "h" MATTERS I DON'T RECOMMEND YOU DO ANY OF THESE THINGS, THOSE ARE ALL VERY DANGEROUS THINGS, ITS BETTER TO DO THIS WITH AN ROCK, AND THAT ACTUALLY THE EXAMPLE WE ARE GOING TO BE CONSIDERING SO LETS JUST THINK ABOUT THIS A LITTLE BIT WE WANT TO FIGURE OUT THIS... AT THE TOP, RIGHT WHEN THE ROCK GETS DROPPED, YOU HAVE AN INITIAL VELOCITY OF "0" AND ONCE AGAIN WE ARE GOING TO USE A CONVENTION HERE THAT POSITIVE VECTOR MEANS UP, AND NEGATIVE VECTOR MEANS DOWN SO WE HAVE INITIAL VELOCITY OF "0", AND AT THE BOTTOM WE GOING TO HAVE SOME FINAL VELOCITY. THAT IS GOING TO BE A NEGATIVE NUMBER OR A NEGATIVE VALUE WE KNOW THAT THE ACCELERATION OF THE GRAVITY FOR AN OBJECT IN FREE FALL NEAR THE SURFACE OF THE EARTH WE GOING TO ASSUME THAT ITS CONSTANT SO OUR CONSTANT ACCELERATION IS GOING TO BE NEGATIVE NINE POINT EIGHT (-9.8) METERS PER SECOND SQUARED SO, THE GIVEN IS "h", THE INITIAL VELOCITY IS ZERO (0 m/s), AND OUR ACCELERATION DUE TO GRAVITY IS 9.8 WE WANT TO FIGURE OUT WHAT OUR FINAL VELOCITY IS GOING TO BE RIGHT BEFORE WE HIT THE GROURD WE GOING TO ASSUME THIS "h" IS GIVEN IN METERS AND THE ANSWER WILL BE IN METERS PER SECOND FOR THIS FINAL VELOCITY SO WE KNOW SOME BASIC THINGS YOU CAN ALWAYS DERIVE THESE MORE INTERESTING QUESTIONS FROM VERY BASIC THING THAT WE KNOW SO WE KNOW THAT THE DISPLACEMENT IS EQUAL TO AVERAGE VELOCITY, TIMES, CHANGE IN TIME AND WE KNOW THAT AVERAGE VELOCITY, IF WE ASSUME ACCELERATION IS CONSTANT, IS THE FINAL VELOCITY + INITIAL VELOCITY OVER TWO. AND THAT OUR CHANGE IN TIME, IS OUR CHANGE IN VELOCITY DIVIDED BY OUR ACCELERATION. JUST TO MAKE SURE YOU UNDERSTAND THIS, IT COMES STRAIGHT FROM THE IDEA THAT CHANGE IN VELOCITY IS JUST ACCELERATION TIMES CHANGE IN TIME DIVIDE BOTH SIDES OF THIS EQUATION BY ACCELERATION YOU GET THIS RIGHT OVER HERE SO THAT IS WHAT OUR DISPLACEMENT IS REMEMBER I WANTED AN EXPRESSION FOR DISPLACEMENT IN TERMS OF THINGS WE KNOW AND THINGS WE WANT TO FIND OUT WELL, FOR THIS EXAMPLE, RIGHT OVER HERE, WE KNOW COUPLE OF THINGS WE KNOW OUR INITIAL VELOCITY IS ZERO SO THIS FIRST EXPRESSION, FOR THE EXAMPLE WE ARE DOING, THE AVERAGE VELOCITY IS GOING TO BE OUR FINAL VELOCITY DIVIDED BY TWO. SINCE OUR INITIAL VELOCITY IS ZERO. CHANGE IN VELOCITY IS THE SAME THING AS FINAL VELOCITY MINUS INITIAL VELOCITY NOW ONCE AGAIN, WE KNOW THAT INITIAL VELOCITY IS ZERO, SO OUR CHANG IN VELOCITY IS SAME AS FINAL VELOCITY SO ONCE AGAIN THIS WILL BE TIMES, INSTEAD OF WRITING CHANGE IN VELOCITY, WE CAN JUST WRITE FINAL VELOCITY BECAUSE WE ARE STARTING AT ZERO INITIAL VELOCITY IS ZERO, TIMES OUR FINAL VELOCITY DIVIDED BY OUR ACCELERATION (ONCE AGAIN) FINAL VELOCITY IS CHANGE IN VELOCITY BECAUSE INITIAL VELOCITY WAS ZERO AND ALL OF THIS IS GOING TO BE OUR DISPLACEMENT AND NOW IT LOOKS LIKE WE HAVE EVERY THING WRITTEN (WHICH) WE KNOW. SO, IF WE MULTIPLY BOTH SIDES OF THIS EXPRESSION BY TWO TIMES OUR ACCELERATION ON THE LEFT HAND SIDE, WE GET TWO TIMES OUR ACCELERATION TIMES OUR DISPLACEMENT IS GOING TO BE EQUAL TO ON THE RIGHT HAND SIDE TWO CANCELS OUT WITH TWO, ACCELERATION CANCELS OUT WITH ACCELERATION IT WILL BE EQUAL TO FINAL VELOCITY SQUARED FINAL VELOCITY SQUARED, FINAL VELOCITY TIME FINAL VELOCITY AND SO, WE CAN JUST SOLVE FOR FINAL VELOCITY HERE. SO WE KNOW WHAT OUR ACCELERATION IS, IT'S -9.8 METERS PER SECOND SQUARED SO THIS IS NEGATIVE 9.8 METERS PER SECOND SQUARED SO IT'S -19.6 METERS PER SECOND SQUARED NOW WHAT OUR DISPLACEMENT IS GOING TO BE DISPLACEMENT OVER THE COURSE OF DROPPING THIS ROCK OFF OF THIS LEDGE OR ROOF SO YOU MIGHT BE TEMPTED TO SAY THAT OUR DISPLACEMENT IS "h" BUT REMEMBER THESE ARE VECTOR QUANTITIES, SO YOU WANT TO MAKE SURE THAT YOU GET DIRECTIONS RIGHT FROM WHERE THE ROCK STARTED TO WHERE IT ENDS, WHAT'S IT DOING? IT'S GOING TO GO A DISTANCE OF "h" DOWNWARDS, AND OUR CONVENTION IS, DOWN IS NEGATIVE SO, IN THIS EXAMPLE, OUR DISPLACEMENT, FROM WHERE ROCK LEAVES YOUR HAND TO WHERE IT HITS THE GROUND THE DISPLACEMENT IS GOING TO BE NEGATIVE "h" IT'G GOING TO TRAVEL A DISTANCE OF "h" BUT ITS GOING TO TRAVEL THAT DISTANCE DOWNWARDS SO THAT'S WHY THIS VECTOR CONVENTION IS VERY IMPORTANT SO OUR DISPLACEMENT OVER HERE IS GOING TO BE "-h" METERS SO, WHEN YOU MULTIPLY THESE TWO THINGS OUT, LUCKY FOR US, THESE NEGATIVE CANCELS OUT AND YOU GET 19.6 h METER SQUARED PER SECOND SQUARED IS EQUAL TO OUR FINAL VELOCITY SQUARED AND NOTICE, WHEN YOU SQUARE SOME THING, YOU LOOSE THE SIGN INFORMATION IF OUR FINAL VELOCITY IS POSITIVE, YOU SQUARED, YOU STILL GET A POSITIVE VALUE AND IF IT WAS NEGATIVE, YOU STILL GET A POSITIVE VALUE BUT REMEMBER IN THIS EXAMPLE WE ARE MOVING DOWNWARDS SO WE WANT THE NEGATIVE VERSION OF THIS SO, TO REALLY FIGURE OUT OUR FINAL VELOCITY, WE TAKE THE NEGATIVE SQUARE ROOT OF BOTH SIDES OF THIS EQUATION SO, IF WE WERE TO TAKE SQUARE ROOT OF BOTH SIDES OF THIS YOU WILL GET, AND I FLIP THEM AROUND, FINAL VELOCITY IS EQUAL TO SQUARE ROOT OF 19.6 h, AND YOU CAN EVEN TAKE THE SQUARE ROOT OF METERS SQUARED PER SECOND SQUARED TREAT THEM LIKE VARIABLE, EVEN THOUGH THEY ARE THE UNITS, AND OUTSIDE OF THE RADICAL SIGH YOU WILL GET METERS PER SECOND AND BE CAREFUL HERE, IF WE JUST TAKE THE PRINCIPLE ROOT, THE PRINCIPLE ROOT HERE IS THE POSITIVE SQUARE ROOT BUT WE KNOW THAT OUR VELOCITY IS GOING TO BE DOWNWARDS HERE BECAUSE THAT IS OUR CONVENTION SO, WE WANT TO MAKE SURE WE GET NEGATIVE SQUARE ROOT. SO, LETS TRY IT OUT WITH SOME NUMBERS WE HAVE ESSENTIALLY SOLVED WHAT WE SET OUT TO SOLVE IN THE BEGINNING OF THIS VIDEO HOW FAST WOULD WE BE FALLING, AS A FUNCTION OF THE HEIGHT LETS TRY IT OUT WITH SOME THING, LETS SAY THAT THE HEIGHT IS 5 METERS WHICH PROBABLY WOULD BE A ONE STORY BUILDING IT WOULD BE ABOUT 15 FEET ABOUT THE ROOF OF A COMMERCIAL BUILDING SO WHAT DO WE GET? IF WE PUT FIVE METERS IN HERE, WE GET 19.6 TIME 5, GIVES US 98, SO ALMOST A HUNDRED AND THEN WE WANT TO TAKE SQUARE ROOT OF THAT SO THE SQUARE ROOT OF 98, IT GIVES US ROUGHLY 9.9 WE WANT THE NEGATIVE SQUARE ROOT OF THAT SO IN THAT SITUATION, WHEN THE HEIGHT IS 5 METERS, IF YOU JUMP OFF A ONE STORY COMMERCIAL BUILDING RIGHT AT THE BOTTOM, IT WILL HAVE A VELOCITY OF -9.9 METERS PER SECOND. SO, NEGATIVE 9.9 METERS PER SECOND I WILL LEAVE IT UP TO YOU AS AN EXERCISE TO HOW FAST THIS IS IN KILOMETERS PER HOUR OR MILES PER HOUR IT'S PRETTY FAST, IT'S NOT SOME THING YOU WANT TO DO BUT YOU CAN NEARLY FIGURE THIS OUT, YOU CAN USE THIS REALLY ANY HEIGHT REASONABLY CLOSE TO THE SURFACE OF THE EARTH AND YOU IGNORE THE AFFECTS OF THE AIR RESISTANCE AT REALLY HIGH HEIGHT, AND WHEN THE OBJECT IS NOT VERY AERODYNAMIC, THEN AIR RESISTANCE WILL MATTER A LOT.