Answer
(a) $2.8~s$
(b) $-29.5~m/s$
Work Step by Step
(a) Let the ground be the origin.
$x_0=45~m$, $x=0$, $v_0=-2.0~m/s$, $a=-g=-9.81~m/s^2$
$x=x_0+v_0t+\frac{1}{2}at^2$
$0=45~m+(-2.0~m/s)t+\frac{1}{2}(-9.81~m/s^2)t^2$
$(4.905~m/s^2)t^2+(2.0~m/s)t-45~m=0$
This is a quadratic equation for $t$.
$t=\frac{-(2.0~m/s)±\sqrt {(2.0~m/s)^2-4(4.905~m/s^2)(-45~m)}}{2(4.905~m/s^2)}$
$t_1=2.8~s$
$t_2=-3.2~s$ (this solution is not valid)
(b) $v=v_0+at$
$v=-2.0~m/s+(-9.81~m/s^2)(2.8~s)=-29.5~m/s$
