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