Answer
(a) $24.8~m/s$
(b) $0.702~s$
Work Step by Step
Let the positive direction be upward.
(a) $t=2.00~s$, $\Delta x=x-x_0=30.0~m$, $a=-g=-9.81~m/s^2$
$x=x_0+v_0t+\frac{1}{2}at^2$
$x-x_0=v_0t+\frac{1}{2}at^2$
$30.0~m=v_0(2.00~s)+\frac{1}{2}(-9.81~m/s^2)(2.00~s)^2$
$30.0~m=v_0(2.00~s)-(19.62~m)$
$v_0=\frac{30.0~m+19.62~m}{2.00~s}=24.8~m/s$
(b) $v_0=24.8~m/s$, $\Delta x=x-x_0=15.0~m$, $a=-g=-9.81~m/s^2$
$x-x_0=v_0t+\frac{1}{2}at^2$
$15.0~m=(24.8~m/s)t+\frac{1}{2}(-9.81~m/s^2)t^2$
$(4.905~m/s^2)t^2-(24.8~m/s)t+15.0~m=0$
This is a quadratic equation for $t$.
$t=\frac{-(-24.8~m/s)±\sqrt {(-24.8~m/s)^2-4(4.905~m/s^2)(15.0~m)}}{2(4.905~m/s^2)}$
$t_1=4.35~s$
$t_2=0.702~s$
The arrow is launched upward. It goes up until it reaches the top then moves back (downward). So, it first reaches a height of $15.0 m$ at $t_2=0.702~s$
