# Anti-derivative with velocity

a ball is thrown upward from the ground with speed of 96 ft/sec. Assuming that acceleration due to gravity is given by a(t)=-32

1-what does the velocity function?

2-what is the position function?

3-how long does it take for ball to return to the ground?

1- $v(t)=-32t+c$

$96=-32(0)+c$ so $c=96$

2-$s(t)=-16t^2+96t+c$

3- when it returns to ground $s(t)=0$ Is this correct and how can I find C?

Yes, it is correct that $s(t)=0$ when it returns to the ground (assuming you're defining zero to be the ground, which is the natural choice here). You are also assuming it is thrown from the ground, so that $s(0)=0.$ Plugging in $t=0$ gives $c=0.$ So to complete the problem, you just need to find the nonzero solution to $0=-16t^2+96t.$