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 1.20m (4.00ft). (This means that he moved upward by 1.20mafter his feet left the floor.) Griffith weighed 881N

What was his speed as he left the floor?

If the time of the part of the jump before his feet left the floor was 0.251s , what was the magnitude of his acceleration (assuming it to be constant) while he was pushing against the floor?

If the time of the part of the jump before his feet left the floor was 0.251s , what was the direction of his acceleration (assuming it to be constant) while he was pushing against the floor?

Use Newton's laws and the results of part B to calculate the force he applied to the ground during his jump.

Explanation / Answer

  Find Time:
It is easier to find the velocity when he hits the ground (which is equivalent to the velocity he left the ground at)
distance = acceleration/2 (t*t) +( v initial) t
v_initial is 0 at his highest point

-1.2 = -4.9 t*t

t = .49 seconds

Making Time Useful, use new equation
(acceleration)(time) + velocity = 0 --> (9.8(.49) = velocity

velocity = 4.8 m/s

the total force (integral of force) is what you are looking for in the second part question. One Newton = 1 m/s * 1 kg, so convert the weight to kg (890/9.8 = 90.8) and multiply by the velocity found above.

4.8*90.8 = 435.8 Newtons

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