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Basketball player Darrell Griffith is on record as
attaining a standing vertical jump of 1.2 m (4 ft).
(This means that he moved upward by 1.2 m after
his feet left the floor.) Griffith weighed 890 N (200
lb). g=9.8 m/s2

1- What is his speed as he leaves the floor?


2- 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?

Sagot :

Explanation:

1.

We use the equation

h = [tex]\frac{gt^2}{2}[/tex], where

h is the height traveled,

g is the acceleration due to gravity and

t is the time taken to reach height h.

We can now calculate t to be

[tex]\sqrt{\frac{2*1.2 m}{9.81 m/s^2} }[/tex]

= 0.495 s

Let v be the initial velocity of the player.

The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.

v = 9.81 m/s^2 x 0.495 s = 4.85 m/s

2.

The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is

4.85 m/s / 0.3 s = 16.2 m/s^2

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