Answer
(a) $19.6~m$
(b) $2.00~s$
Work Step by Step
$h$ -> height
Let the positive direction be downward.
(a) First, let's find the velocity of the ball after it has covered the first quarter of the distance to the ground ($\frac{1}{4}h$).
$a=g$, $v_0=0$, $\Delta x=\frac{1}{4}h$
$v^2=v_0^2+2a\Delta x$
$v^2=0^2+2g\frac{1}{4}h=\frac{1}{2}gh$
$v=±\sqrt {\frac{1}{2}gh}$
$v=\sqrt {\frac{1}{2}gh}$. The ball is moving downward (positive direction).
Now, in the remaining three quarters of the distance to the ground, $\sqrt {\frac{1}{2}gh}$ is the initial velocity:
$\Delta x=\frac{3}{4}h$, $a=g=9.81~m/s^2$, $t=1.00~s$, $v_0=\sqrt {\frac{1}{2}gh}$
$x=x_0+v_0t+\frac{1}{2}at^2$
$x-x_0=(\sqrt {\frac{1}{2}gh})t+\frac{1}{2}gt^2$
$\Delta x=(\sqrt {\frac{1}{2}gh})t+\frac{1}{2}gt^2$
$\frac{3}{4}h=(\sqrt {\frac{1}{2}gh})t+\frac{1}{2}gt^2$
$-(\sqrt {\frac{1}{2}gh})t=\frac{1}{2}gt^2-\frac{3}{4}h$. Multiply both sides by -1.
$(\sqrt {\frac{1}{2}gh})t=\frac{3}{4}h-\frac{1}{2}gt^2$. Square both sides.
(Restriction: $\frac{3}{4}h-\frac{1}{2}gt^2\geq0$ -> $\frac{3}{4}h\geq\frac{1}{2}gt^2$ -> $h\geq\frac{2}{3}gt^2$)
$\frac{1}{2}ght^2=\frac{1}{4}g^2t^4-\frac{3}{4}ght^2+\frac{9}{16}h^2$
$0=\frac{1}{4}g^2t^4-\frac{5}{4}ght^2+\frac{9}{16}h^2$. Multiply both sides by 16.
$0=4g^2t^4-20ght^2+9h^2$ or
$(9)h^2+(-20gt^2)h+(4g^2t^4)=0$
This is a quadratic equation for $h$.
$h=\frac{-(-20gt^2)±\sqrt {(-20gt^2)^2-4(9)(4g^2t^4)}}{2(9)}$
$h=\frac{20gt^2±\sqrt {400g^2h^2t^4-144g^2t^4}}{18}$
$h=\frac{20gt^2±\sqrt {256g^2h^2t^4}}{18}=\frac{20gt^2±16gt^2}{18}$
$h_1=\frac{36gt^2}{18}=2gt^2$.
Notice that: $h_1\geq\frac{2}{3}gt^2$. Ok.
$h_1=2(9.81~m/s^2)(1.00~s)^2=19.62~m\approx19.6~m$
$h_2=\frac{4gt^2}{18}=\frac{2}{9}gt^2$.
Notice that: $h_2\lt\frac{2}{3}gt^2$. Not valid.
We can conclude that: $h=19.6~m$
(b) $\Delta x=19.62~m$, $a=g=9.81~m/s^2$, $v_0=0$
$x=x_0+v_0t+\frac{1}{2}at^2$
$x-x_0=v_0t+\frac{1}{2}at^2$
$\Delta x=v_0t+\frac{1}{2}at^2$
$19.62~m=0t+\frac{1}{2}(9.81~m/s^2)t^2$
$39.24~m=(9.81~m/s^2)t^2$
$t=±\sqrt {\frac{39.24~m}{9.81~m/s^2}}$. The valid soution is:
$t=2.00~s$
