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 ).

(a) What is his speed as he leaves the floor?
answer: v = 4.9m/s

(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?
answer: a = m/s^2

(c) What is its direction?
answer: upward

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

Explanation / Answer

a)let his speed be v so v^2 = u^2 + 2as u^2 = 2*9.8*1.2 u = sqrt(2*9.8*1.2) = 4.84 m/s = 4.9 m/s b) Change in momentum = m.v = 90.81*4.9 = 445 time = .3 sec Force = 1483.333 N Acceleration = 16.33 m/s^2 c) Answer is directed upwards d)Average force he applied to the ground to jump = mg+1483.33 = 890+1483.33 = 2373.33 N

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