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

where v is final velocity which is zero at max height and u is it initial
hence


now we can find time in the 15 cm ascent


using quadratic formula

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

where s=0.76-0.15
as the player has traveled the above distance to reach 15cm to the bottom


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

t=0.0409
I believe the answer is #4. u can always ask google if u believe that's the wrong answer :)
Here in order to find out the distance between two planes after 3 hours can be calculated by the concept of relative velocity

here
speed of first plane is 700 mi/h at 31.3 degree


speed of second plane is 570 mi/h at 134 degree


now the relative velocity is given as


now the distance between them is given as



so the magnitude of the distance is given as

miles
so the distance between them is 2985.6 miles
Its d all the above your welcome