Basketball player Darrell Griffith is on record as attaining a standing vertical
ID: 249311 • Letter: B
Question
Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.20 m (4.00 ft ). (This means that he moved upward by 1.20 m after his feet left the floor.) Griffith weighed 891 N .(A)What was his speed as he left the floor?(B)If the time of the part of the jump before his feet left the floor was 0.220 s , what was the magnitude of his acceleration (assuming it to be constant) while he was pushing against the floor?(C)If the time of the part of the jump before his feet left the floor was 0.220 s , what was the direction of his acceleration (assuming it to be constant) while he was pushing against the floor?(D)Draw a free-body diagram of Griffith during the jump.(E)Use Newton's laws and the results of part B to calculate the force he applied to the ground during his jump.
Explanation / Answer
Given that the weight of the person is W = 891 and maximum height is H = 1.20 m
(a) From the equation of motion final velocityis V2- U2= 2as
0 - U2 = -2gH
U = 2*g*H
= 2*9.8m/s2*1.20m
=4.85 m/s
(b) Time of contact is t = 0.220s
From the equation of motion U = v + at
since initial velocity v = 0 then U = at
a= U /t
=4.85 m/s / 0.220s
= 22.045 m/s2
This is the accelaration of theperson
(c) The force is upwards
(e) Force exerted on the ground is F = mg + ma
=W + (W/g)a
=891 N + (891N/9.8m/s2)(22.045 m/s2)
= 2895.30 N
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