We're asked to find the projectile's distance from the launch point when it reaches its maximum height. We therefore need to find its #x#- and #y#-position when it's at this point.
Let's first, for reference later, find its initial velocity components:
#v_(0x) = v_0cosalpha = (16"m"/"s")cos(pi/8) = 14.8"m"/"s"#
#v_(0y) = v_0sinalpha = (16"m"/"s")sin(pi/8) = 6.12"m"/"s"#
The point at which the particle is at its highest point is when its instantaneous #y#-velocity is equal to #0#. We can find this #y#-position by the equation
#(v_y)^2 = (v_(0y))^2 + 2a_y(Deltay)#
The acceleration is #-g#, which is #-9.8"m"/("s"^2)#. The velocity #v_y# is #0#, and the initial #y#-velocity was just found to be #6.12"m"/"s"#. Plugging in these values, and solving for the height #Deltay#, we have
#(0)^2 = (6.12"m"/"s")^2 + 2(-9.8"m"/("s"^2))(Deltay)#
#19.6"m"/("s"^2)(Deltay) = 37.5("m"^2)/("s"^2)#
#Deltay = color(red)(1.91# #color(red)("m"#
Now, with this value, let's calculate the time #t# when this occurs, so that we can then find out how far in the #x#-direction it is. We can use the formula
#v_y = v_(0y) + a_yt#
We already know everything here to solve for the time, so let's solve this for #t#:
#0 = 6.12"m"/"s" + (-9.8"m"/("s"^2))t#
#(9.8"m"/("s"^2))t = 6.12"m"/"s"#
#t = 0.625# #"s"#
Now, with this time we can use the formula
#Deltax = v_(0x)t#
to calculate the horizontal distance. We calculated the initial #x#-velocity earlier to be #14.8"m"/"s"#, so
#Deltax = (14.8"m"/cancel("s"))(0.625cancel("s")) = color(green)(9.24# #color(green)("m")#
Lastly, we can calculate the total distance #r# from the launch point using the distance formula:
#r = sqrt((Deltax)^2 + (Deltay)^2) = sqrt((9.24"m")^2 + (1.91"m")^2)#
#= color(blue)(9.59# #color(blue)("m"#
Therefore, when the projectile is at its maximum height, it is #color(blue)(9.59# meters from its launch point.
