Answer
(a) $0.61~s$
(b) $0.45~m$
Work Step by Step
Let the positive direction be upward and the origin at the bottom of the window.
Let the initial time be the instant the ball goes through the bottom of the window.
$x_0=0$, $x=1.05~m$, $a=-g=-9.81~m/s^2$, $t=0.25~s$
$x=x_0+v_0t+\frac{1}{2}at^2$
$1.05~m=0+v_0(0.25~s)+\frac{1}{2}(-9.81~m/s^2)(0.25~s)^2$
$-(0.25~s)v_0=-1.05~m-\frac{1}{2}(9.81~m/s^2)(0.25~s)^2$
$v_0=\frac{-1.05~m-\frac{1}{2}(9.81~m/s^2)(0.25~s)^2}{-(0.25~s)}=5.42625~m/s$
That is the velocity of the ball in the moment it goes through the bottom of the window. Now, let's find its velocity in the moment it goes through the top of the window:
$v=v_0+at$
$v=5.42625~m/s+(-9.81~m/s^2)(0.25~s)=2.97375~m/s$
(a) Let the initial time be the instant the ball goes through the top of the window, moving upward.
When the ball reappears, in the top of the window, it is moving downward, but the magnitude of the velocity is the same as the magnitude of the velocity in the the instant the ball was going through the top of the window, moving upward. Then:
$v_0=2.97375~m/s$, $v=-v_0=-2.97375~m/s$, $a=-g=-9.81~m/s^2$
$v=v_0+at$
$-2.97375~m/s=2.97375~m/s+(-9.81~m/s^2)t$
$(9.81~m/s^2)t=2.97375~m/s+2.97375~m/s$
$t=\frac{5.9475}{9.81}=0.61~s$
(b) When the ball reaches the greastest height its velocity is zero.
$v_0=2.97375~m/s$, $v=0$, $a=-g=-9.81~m/s^2$
$v^2=v_0^2+2a\Delta x$
$0^2=(2.97375~m/s)^2+2(-9.81~m/s^2)\Delta x$
$(19.62~m/s^2)\Delta x=2.97375~m/s$
$\Delta x=\frac{(2.97375~m/s)^2}{19.62~m/s^2}=0.45~m$
