Answer
(a) $x=11~m$
(b) $v=15~m/s$
(c) $t=2.1~s$
Work Step by Step
Let the origin be on the railroad bridge with the positive direction downward.
$x_0=0$, $v_0=0$, $a=g=9.81~m/s^2$, $t=1.5~s$
(a)
$x=x_0+v_0t+\frac{1}{2}at^2$
$x=0+0t+\frac{1}{2}(9.81~m/s^2)(1.5~s)^2$
$x=11~m$
(b)
$v=v_0+at$
$v=0+(9.81~m/s^2)(1.5~s)$
$v=15~m/s$
(c)
Now, $x=2\times(11~m)=22~m$
$x=x_0+v_0t+\frac{1}{2}at^2$
$22~m=0+0t+\frac{1}{2}(9.81~m/s^2)t^2$
$t=\sqrt {\frac{2\times(22~m)}{9.81~m/s^2}}$
$t=2.1~s$
