A ball is thrown vertically upward with an initial velocity of 96 feet per second.The...

We have a ball is thrown vertically upward with an initial velocity of 96 feet per second.The distance s (in feet) of the ball from the ground after t seconds is ${\displaystyle s\left(t\right)=96t-16t^2}$. 

a) we have ${\displaystyle s\left(t\right)=96\left(t\right)-16t^2=0}$ 
since the distance from the ground is s(t) so we have 
${\displaystyle s\left(t\right)=96t-16t^2=0}$ 
${\displaystyle t(96-16t)=0}$ 
${\displaystyle t=0}$ or ${\displaystyle t=6}$ 
the ball is on the ground at t=0 seconds and t=6 seconds when the ball is thrown and when the ball lands. 
So, the ball strikes the ground after 6 seconds.

b) when is ${\displaystyle s=128}$ so putting this value in the function then we have 
${\displaystyle 128=96t-16t^2}$ 
${\displaystyle t^2-6t+8=0}$ 
${\displaystyle (t-4)(t-2)=0}$ 
that is between 2 seconds and 4 seconds.