We need to find the maximum range of the projectile, assuming its launch and landing heights are equal.
In terms of physics, we need to find the distance #Deltax# (#x# the horizontal component) of the projectile when the change in vertical position #Deltay# is #0#.
The initial #y#-velocity #v_(0y)# of the projectile is
#v_(0y) = v_0sinalpha = 3"m"/"s"sin(pi/8) = 1.15 "m"/"s"#
and, since we'll need this later, the initial #x#-velocity #v_(0x)# is
#v_(0x) = v_0cosalpha = 3"m"/"s"cos(pi/8) = 2.77 "m"/"s"#
We can use the equation
#Deltay = v_(0y)t - 1/2g t^2#
to find the time #t# when the change in height #Deltay# is zero.
#0 = v_(0y)t - 1/2g t^2#
#1/2g t^2 = v_(0y)t#
We can see that one of these times is #t = 0#, because at this time, the motion hasn't started yet, so intuitively it hasn't changed it's height yet.
(There will always be two calculated times where the height #Deltay# (or #h#) is equal to #0# (or any value for that matter), because we're in essence using the above equation which can be solved using the quadratic equation, which always yields two answers. Picture it like this: if you throw a ball upward and its maximum height is #15"m"# above the ground, there will be two times when its height is equal to, say, #10"m"#: one time on the way up, and the other time on the way down.)
We can divide both sides by the variable #t# to yield
#1/2g t = v_(0y)#
So,
#t = (2v_(0y))/g#
Plugging in known variables, we have
#t = (2(1.15"m"/"s"))/(9.80"m"/"s"^2) = 0.235 "s"#
Lastly, to find the horizontal range #Deltax# (or sometimes #R#), we'll use the equation
#Deltax = v_(0x)t#
Plugging in known variables yields
#Deltax = (2.77"m"/"s")(0.235"s") = color(red)(0.651"m"#
If you want to round this to one significant figure, the answer would technically be #0.7 "m"#. Ideally, measurements like those in this problem should be given with more than one significant figure to limit uncertainty.
