A young lady is jumping rope; she always jumps so that both feet leave the groun

Question

A young lady is jumping rope; she always jumps so that both feet leave the ground at the same instant of time. Each of her jumps reaches a height of 14.5 cm off of the ground. (This means that she moves upward by 14.5 cm after her feet leave the ground.) The weight of the young lady is 429 N .

Part A: What is her speed each time she leaves the ground? (Answer: v=..........m/s)

Part B: Assuming that her landings are a time-wise mirror image of her jumps (i.e. a movie of one of her landings, run backwards, would be indistinguishable from a jump), for how long is she in the air each time she jumps? (Answer in Seconds)

Part C: Each time that she is on the ground, she is first landing and then jumping (pushing off). At one instant of time while she is on the ground, she is momentarily at rest (when she is in the lowest point of her crouch); her jump begins from this instant when she is momentarily at rest. If the elapsed time of the part of her jump before her feet leave the floor (i.e. from momentarily at rest until toes leaving the floor) is always 0.270 s , what is the magnitude of her average acceleration while she is pushing against the floor? (Answer: a=........ m/s^2)

Part D: What is the direction of the acceleration described in Part C? (Upward or Downward)

Part E: Assuming that the young lady's acceleration is constant while she is touching the ground, then what is the magnitude of the net force acting on the young lady while she is touching the ground? (Answer in Newtons)

Part F: Consider any and all downward forces that might be acting on the young lady while she is touching the ground. What is the magnitude of the sum of all these downward forces (there may possibly be only one downward force)? (Answer in Newtons)

Part G: Consider any and all upward forces that might be acting on the young lady while she is touching the ground. Assuming that the young lady's acceleration is constant while she is touching the ground, then what is the magnitude of the sum of all these upward forces (there may possibly be only one upward force)? (Answer in Newtons)

Part H: Assuming that the young lady's acceleration is constant while she is touching the ground, then calculate the size of the force she is applying to the ground while she is touching the ground. (Answer: F=........N)

Part I: The direction of the force described in Part H is: (Choose one):

1- downward   2- the force is zero, so no direction   3- upward

Part J: What is the force which is responsible for launching the young lady into the air? (Choose one)

1- the weight of the young lady

2- the contact force on the ground by the young lady

3- the contact force on the young lady by the ground

4- the force of gravity on the Earth by the young lady

Explanation / Answer

a)v=sqrt(2gh)=sqrt(2*9.81*0.145)=1.69m/s

b)T=2v/g=0.344s

c)a=(vf-vi)/t=1.69/0.27=6.26m/s^2

d)upward.

e)F=ma=274N

f)F=W=429N

g)F upward=W+Net upward force=703N

h)F=703N (by newton's third law)

i) downward

j)- the contact force on the young lady by the ground

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