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 1.2 (4 ). (This means that he moved upward by 1.2 after his feet left the floor.) Griffith weighed 890 (200 ).

Part B
If the time of the part of the jump before his feet left the floor was 0.300 , what was the magnitude of his average acceleration while he was pushing against the floor?
Express your answer using two significant figures.
=16m/s2

Part C
Use Newton's laws and the results of part (B) to calculate the average force he applied to the ground.

* What I need help calculating is part C, The answer to part B is 16 m/s2. All help is appreciated. Thank You!!!!

Explanation / Answer

a(y)=-9.8 m/s^2 y-y(o)=1.2 m V(y)^2=2*a(y)*(y-y(o)) V(y)=4.89 m/s a(av)=V(y)/t=4.89/0.3 = 16.2 m/s Now Your part solution: mass of player = 890/9.8 = 90.8 kg F(av)=mg+m*a(av) =90.8(9.8+16.2) =2360 N

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.