Answer:
The time for final 15 cm of the jump equals 0.1423 seconds.
Explanation:
The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as
[tex]v^2=u^2+2as[/tex]
where
'v' is the final velocity of the player
'u' is the initial velocity of the player
'a' is acceleration due to gravity
's' is the height the player jumps
Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get
[tex]0^2=u^2-2\times 9.81\times 0.76\\\\\therefore u=\sqrt{2\times 9.81\times 0.76}=3.86m/s[/tex]
Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as
[tex]v^{2}=3.86^2-2\times 9.81\times 0.66\\\\\therefore v=\sqrt{3.86^2-2\times 9.81\times 0.66}=1.3966m/s[/tex]
Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as
[tex]v=u+at[/tex]
where symbols have the usual meaning
Applying the given values we get
[tex]t=\frac{v-u}{g}\\\\t=\frac{0-1.3966}{-9.81}=0.1423seconds[/tex]
