We're asked to find the horizontal range of a projectile given its initial speed and launch angle.
To do this, we can first find the time #t# when it has a height of #0#, using the equation
#Deltay = v_(0y)t - 1/2g t^2#
The initial #y#-velocity #v_(0y)# is
#v_(0y) = v_0sinalpha_0 = (11color(white)(l)"m/s")sin(pi/4) = 7.78# #"m/s"#
The change in height #Deltay# is #0#, because we're trying to find the time #t# at this height. Plugging in known values, we have
#0 = (7.78color(white)(l)"m/s")t - 1/2(9.81color(white)(l)"m/s"^2)t^2#
#(4.905color(white)(l)"m/s"^2)t^2 = (7.78color(white)(l)"m/s")t#
#(4.905color(white)(l)"m/s"^2)t = 7.78color(white)(l)"m/s"#
#t = color(red)(1.59# #color(red)("s"#
We can now use the equation
#Deltax = v_(0x)t#
to find the horizontal range, #Deltax#
The initial #x#-velocity #v_(0x)# is
#v_(0x) = v_0cosalpha_0 = (11color(white)(l)"m/s")cos(pi/4) = 7.78# #"m/s"#
We then have:
#Deltax = (7.78"m"/(cancel("s")))(1.59cancel("s")) = color(blue)(12.3# #color(blue)("m"#
