Basketball player Darrell Griffith is on record as attaining a standing vertical

Question

Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2m (4ft ). (This means that he moved upward by 1.2m after his feet left the floor.) Griffith weighed 890 N(200 lb).
a) What is his speed as he leaves the floor?
b) If the time of the part of the jump before his feet left the floor was 0.300s , what was the magnitude of his average acceleration while he was pushing against the floor?
c)What is its direction?
upward
downward
d) Use Newton's laws and the results of part (B) to calculate the average force he applied to the ground.

Explanation / Answer

the answer to A & B use the work energy equation let the speed be v then 1/2mv^2-mgh=0so v=sqrt2gh=sqrt(2*10*1.2)=sqrt24m/s=4.8m/… during .300s he is acted upon by an impulsive force which causes change in momentum . we know f*t=mv-mu so the force(impulsive)is 89*4.8units again p=mf(p=force and f=acceleration) so the acceleration is (89*4.8/.300/89)m/s^2=16m/s^2 so the results are 4.8m/s and 16m/s^2 the acceleration is in upward direction.

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