- #1
JSpadafore
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Hello! This problem has been driving me crazy all week long, and now I am down to the last few hours before the assignment is due. Any help would be greatly appreciated!
An average person can reach a maximum height of about 60cm when jumping straight up from a crouched position. During the jump itself, the person's body from the knees up typically rises a distance of around 50cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump.
Part A required me to find the initial speed a person would need to reach a maximum height of 60cm, for which I correctly answered 3.4m/s.
Part B is where I find myself stuck. It asks, "In terms of this jumper's weight (w), what force does the ground exert on him or her during the jump?"The answer will read:
F= ___ w.
ƩFy=may
w=mg
...
Honestly, I've just been trying to figure out how I am supposed to get anywhere without first knowing the mass of the person jumping. I suspect it has something to do with the initial speed I had calculated for Part A being related to an opposite force, coming from the ground going to the person jumping.
Homework Statement
An average person can reach a maximum height of about 60cm when jumping straight up from a crouched position. During the jump itself, the person's body from the knees up typically rises a distance of around 50cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump.
Part A required me to find the initial speed a person would need to reach a maximum height of 60cm, for which I correctly answered 3.4m/s.
Part B is where I find myself stuck. It asks, "In terms of this jumper's weight (w), what force does the ground exert on him or her during the jump?"The answer will read:
F= ___ w.
Homework Equations
ƩFy=may
w=mg
...
The Attempt at a Solution
Honestly, I've just been trying to figure out how I am supposed to get anywhere without first knowing the mass of the person jumping. I suspect it has something to do with the initial speed I had calculated for Part A being related to an opposite force, coming from the ground going to the person jumping.
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