Respuesta :
Answer: Part(a)=0.041 secs, Part(b)=0.041 secs
Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction
now we know that
a=-9.81 m/s^2 ( negative because it is pulling the player downwards)
we also know that
s=76 cm= 0.76 m ( maximum s)
using kinetic equation
[tex]v^2=u^2+2as[/tex]
where v is final velocity which is zero at max height and u is it initial
hence
[tex]u^2=-2(-9.81)*0.76[/tex]
[tex]u=3.8615 m/s\\[/tex]
now we can find time in the 15 cm ascent
[tex]s=ut+0.5at^2[/tex]
[tex]0.15=3.861*t+0.5*9.81t^2\\[/tex]
using quadratic formula
[tex]t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}[/tex]
t=0.0409 sec
the answer for the part b will be the same
To find the answer for the part b we can find the velocity at 15 cm height similarly using
[tex]v^2=u^2+2as[/tex]
where s=0.76-0.15
as the player has traveled the above distance to reach 15cm to the bottom
[tex]v^2=0^2 +2*(9.81)*(0.76-0.15)[/tex]
[tex]v=3.4595[/tex]
when the player reaches the bottom it has the same velocity with which it started which is 3.861
hence the time required to reach the bottom 15cm is
[tex]t=\frac{3.861-3.4595}{9.81}[/tex]
t=0.0409
