Great question. The way we envision a jump is the following: the compressed legs extend upward (accelerate at 29.4 m/s^2) until they are straight, at which point the legs are not continuing to provide any upward acceleration, but the person has reached a speed of 7.35 m/s. If you started your camera from this point, it would appear as if a straight-legged person was launched upward at 7.35 m/s and is now undergoing gravitational acceleration of -9.8 m/s^2. The "height" of their jump is the distance between the ground and their foot. Hope this helps. Cheers, Dr. A
a (up) is 29.4 m/s^2 and a(land) is 58.8 m/s^2 which is double the acceleration in air , wouldn't he just fly off into outer space. I think a (land) should be negative i.e., -58.8 m/s^2.
a (up) is 29.4 m/s^2 and a(land) is 58.8 m/s^2 which is double the acceleration in air , wouldn't he just fly off into outer space. I think a (land) should be negative i.e., -58.8 m/s^2.
hello sir, YOUR GRAPH IS COMPLETELY WRONG!! the basic idea of jump is that body gain net positive force, which help them to accelerate away from the ground. while their are many types of forces; which for the ease of understanding can be classified as pull or push in the explanation, forces are vector quantity and therefore they have direction, for now we are only taking one dimension in account which is the y direction. for this experiment we can form three kinds of graph to explain the concept 1) force acting on the body/ force exerted by the ground 2)force exerted by the body/ force acting on the ground 3) the net force on body but in your video you started by a graph saying "the person have some force on the ground" which i believe you were trying to say that the force of gravity acting on the body or the force of body acting on the gravity which is a PULL FORCE, then you said they exert a positive force which you draw in the same direction as the initial force which is completely wrong it should always be in opposite direction and the force in mid air should always be same as the initial force. for any question regarding this ply reply
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Can you make lab experiment with this problem? :)
While solving how high is the jump, why we didn't take a(y) i.e. 29.4 m/sq.sec and Vf i.e. 7.35 m/sec instead of zero?
Great question. The way we envision a jump is the following: the compressed legs extend upward (accelerate at 29.4 m/s^2) until they are straight, at which point the legs are not continuing to provide any upward acceleration, but the person has reached a speed of 7.35 m/s. If you started your camera from this point, it would appear as if a straight-legged person was launched upward at 7.35 m/s and is now undergoing gravitational acceleration of -9.8 m/s^2. The "height" of their jump is the distance between the ground and their foot.
Hope this helps.
Cheers,
Dr. A
a (up) is 29.4 m/s^2 and a(land) is 58.8 m/s^2 which is double the acceleration in air , wouldn't he just fly off into outer space. I think a (land) should be negative i.e., -58.8 m/s^2.
a (up) is 29.4 m/s^2 and a(land) is 58.8 m/s^2 which is double the acceleration in air , wouldn't he just fly off into outer space. I think a (land) should be negative i.e., -58.8 m/s^2.
the graph isnt understanding,as a person has to jump he has to go down which may effect
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hello sir,
YOUR GRAPH IS COMPLETELY WRONG!!
the basic idea of jump is that body gain net positive force, which help them to accelerate away from the ground.
while their are many types of forces; which for the ease of understanding can be classified as pull or push in the explanation,
forces are vector quantity and therefore they have direction, for now we are only taking one dimension in account which is the y direction.
for this experiment we can form three kinds of graph to explain the concept
1) force acting on the body/ force exerted by the ground 2)force exerted by the body/ force acting on the ground 3) the net force on body
but in your video you started by a graph saying "the person have some force on the ground" which i believe you were trying to say that the force of gravity acting on the body or the force of body acting on the gravity which is a PULL FORCE, then you said they exert a positive force which you draw in the same direction as the initial force which is completely wrong it should always be in opposite direction and the force in mid air should always be same as the initial force.
for any question regarding this ply reply