The velocity is the rate of change of the object's position:
\(s(t)=159-16t^2\implies v(t)=s'(t)=-32t\)
Acceleration is the rate of change of velocity:
\(a(t)=v'(t)=s''(t)=-32\)
The time it takes for the ball to hit the ground is the time \(t\) for which \(s(t)=0\), since \(s(t)\) describes the position of the ball above the ground.
\(159-16t^2=0\implies 159=16t^2\implies t=\sqrt{\dfrac{159}{16}}\approx3.15\)
(in seconds)
The object's velocity at this time can be found by plugging in this value of \(t\) in to the velocity function.
\(v(3.15)\approx-32(3.15)\approx100.88\)
(in feet per second)
