So like the first part,
Vf = Vi + a*t
This time we solve for t
t = (Vf - Vi) / a
Vf = 0 m/s
Vi = V sin 45
a = -9.81 m/s^2
t = (Vf - V sin 45)/a
t = -V sin 45 / a
The total flight time is actually twice this time, lets call that T, T = 2t = -2Vsin45/a
we could actually stop here and calculate the time in seconds, but there is no need, really, lets keep it as an equation.
Now for the horizontal distance, we use Xf = Xi + Vx* T
Xi = 0 m
Xf = V cos 45 * T
Substitute in our equation for time.
Xf = V cos 45 * (-2Vsin45/a)
Xf = -2 * (V^2) (sin 45 * cos 45)/a
sin 45 * cos 45 = 0.5
Xf = - (V^2)/a ........[eq 2]
Xf = - (19.62 m/s)^2 / -9.812 m/s^2
Vf = 39.24 m
Mine probably differs from theirs a little because I don't have as much round off error, because I didn't stop and calculate out a number at each stage, and then rounding it off and entering it back into the next equation. By keeping it all as variables until the very end, you can see that many of the variables fell out and let a fairly easy final equation;.
Hmm, I wonder what would have happened if we didn't even stop to calculate out the initial velocity in the first part of the problem, and kept it ALL as variables.
Let's call the time that we are given in the problem as (Tv) for vertical time, and call (tv) the half of Tv the we calculated. (In a perfect world, we should be able to come up with an equation where the only variable is the one that they give us, TV = 4 sec, and maybe acceleration of gravity.)
Lets start back on [eq 1].
Vi = -a*tv and Tv = 2tv
Vi = -a*Tv/2
so in our later equation, V = -a * Tv /2
Now lets plug that into [eq 2]
Xf = - (V^2)/a
Xf = - (( -a * Tv /2)^2)/a = a^2 * Tv ^2 / 4 * a
Xf = a * Tv^2/4
Xf = 9.81 m/s^2 * (4 sec)^2 /4
Xf = 39.24 m
Oh yeah, now that;s the simplest formula, Xf = a * Tv^2 /4
And that's why I like to carry the variables all the way through, saves all kinds of calculator time and round-off error,.