Answer
a. $v(0)=96$ ft/sec.
b. $t=3sec$, $s(3)=256$ ft
c. $v(7)=-128$ ft/sec (downward).
Work Step by Step
Given the equation of motion $s(t)=-16t^2+96t+112$ ft, we have $v(t)=s'=-32t+96$ ft/sec and $a=v'=-32ft/s^2$
a. When $t=0$. $v(0)=-32(0)+96=96$ ft/sec.
b. At its maximum height, we have $v(t)=0$ or $-32t+96=0$ which gives $t=3sec$ and $s(3)=-16(3)^2+96(3)+112=256$ ft
c. When $s=0$, we have $-16t^2+96t+112=0$ or $t^2-6t-7=0$ which gives $t=-1, 7$. Discard the negative answer; we have $t=7sec$ and $v(7)=-32(7)+96=-128$ ft/sec (downward).
