We're asked to find the distance from the starting point a projectile is when it reaches its maximum height, given its initial velocity.

To do this, we need to find the vertical *and* horizontal components of the position at this point. For reference during our work, the components of the initial velocity are

#v_(0x) = (8color(white)(l)"m/s")cos((5pi)/12) = 2.07# #"m/s"#

#v_(0y) = (8color(white)(l)"m/s")sin((5pi)/12) = 7.73# #"m/s"#

**Vertical Position**

To find the height #Deltay# when the projectile is at its maximum height, we can use the equation

#(v_y)^2 = (v_(0y))^2 + 2a_y(Deltay)#

The acceleration #a_y# is #-g#. At its maximum height, the instantaneous velocity #v_y = 0#, so we have

#0 = (7.73color(white)(l)"m/s")^2 - 2(9.81color(white)(l)"m/s"^2)(Deltay)#

#2(9.81color(white)(l)"m/s"^2)(Deltay) = 59.7# #"m"^2"/s"^2#

#Deltay = (59.7color(white)(l)"m"^2"/s"^2)/(2(9.81color(white)(l)"m/s"^2)) = color(red)(3.04# #color(red)("m"#

The time #t# when this occurs is given by

#v_y = v_(0y) - g t#

#t = (v_y - v_(0y))/(-g) = (0 - 7.73color(white)(l)"m/s")/(-9.81color(white)(l)"m/s"^2) = 0.788# #"s"#

We'll use this in calculating the horizontal component below.

**Horizontal Position**

The horizontal position #Deltax# is given by

#Deltax = v_(0x)t#

We found #t# above, so plugging in known values, we have

#Deltax = (2.07color(white)(l)"m/s")(0.788color(white)(l)"s") = color(green)(1.63# #color(green)("m"#

**Distance**

The distance #r# from the starting point is given by

#r = sqrt((Deltax)^2 + (Deltay)^2)#

Plugging in:

#r = sqrt((color(green)(1.63)color(white)(l)color(green)("m"))^2 + (color(red)(3.04color(white)(l)color(red)("m")))^2) = color(blue)(3.45# #color(blue)("m"#