Answer
(a) $\frac{1}{2}gt^2$
(b) $gt$
(c) $13.4~m$
(d) $16.2~m/s$
Work Step by Step
Let the positive direction be upward with origin at the ground.
(a) At maximum height $v=0$
$v=0$, $x_0=0$, $a=-g$
$v=v_0+at$
$0=v_0-gt$
$v_0=gt$. Part (b)
$x=x_0+v_0t+\frac{1}{2}at^2$
$x=0+(gt)t+\frac{1}{2}(-g)t^2$
$x=gt^2-\frac{1}{2}gt^2=\frac{1}{2}gt^2$
(b) $v_0=gt$
(c) $x=\frac{1}{2}gt^2=\frac{1}{2}(9.81~m/s^2)(1.65~s)^2=13.4~m$
(d) $v_0=gt=(9.81~m/s^2)(1.65~s)=16.2~m/s$
