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"#
