First convert the launch angle to degrees and that is
#(cancel(\pi)/(12) cancel(rad))((360°)/(2cancel(\pi)cancel(rad)))=15°#
Much more, it is more convenient to assume that the launch was initially done at the origin.
Deriving the equation requires the x and y components of velocity.
Where
#(1) V_(x)=V_(0x)+cancel(at)=>V_x = V_0costheta#
#(2) V_(y)=V_(0y)+at=> V_y = V_0 sintheta - (g)(t)#
Note that there is no acceleration along the x-direction which explains why #at# in equation #(1)# is cancelled.
Also,
From the analogy of the equation #x=x_0+V_(0x)t+(at^2)/2#,
#(3)x=cancel(x_0)+V_(0x)t+cancel((at^2)/(2))#
#(4)y=cancel(y_0)+V_(0y)t-(g(t^2))/(2)#
In equation #(3)#, #(at^2)/(2)# is cancelled because again, there is no acceleration along the x-direction. And in both equations, #x_0# and #y_0# are both cancelled since the problem is assumed/assigned to initially progress at the origin.
To facilitate the determination of the maximum range of the trajectory, the total flight time must be derived. The total flight time is twice the time it takes for an object to reach it maximum flight height (remember, the maximum flight height is located at the trajectory's/parabola's symmetry point. i.e. it takes #t/2# to reach the first half, and another #t/2# to reach the other half for a total #t# flight time.).
using equation #(2)#, the time needed to reach the object's maximum flight height (or the first half of the trajectory). It can be derived knowing that the y-velocity becomes 0 when it reaches its maximum flight height.
Therefore, #0=V_0sintheta-((g)(t))#
#(5)=>t = (V_0sintheta)/(g)#
Equation #(5)# is again the time needed to reach the object's maximum flight height (or the first half of the trajectory). Therefore, multiplying equation #(5)# by 2 is equal to the total flight time.
#(6)=>t=(2V_0sintheta)/g#
Substituting equation #(6)# into equation #(3)#
#x=V_(0x)t=(V_(0x))((2V_0sintheta)/g)=(V_0costheta)((2V_0sintheta)/g)#
#x=R=(2V_0^2sinthetacostheta)/g#
