Question

Projectile launched from a cannon at an elevation angle of and velocity vl.

Answer 1

As you described, we substitute y=0y=0and x=Rx=R into the trajectory equation:

0=H+Rtanθ−R2g2u2sec2θ.(1)(1)0=H+Rtan⁡θ−R2g2u2sec2⁡θ.

Then, differentiating with respect to θθ and setting dRdθ=0dRdθ=0:

0=Rmaxsec2θ−R2maxg2u22sec2θtanθ,0=Rmaxsec2⁡θ−Rmax2g2u22sec2⁡θtan⁡θ,

which simplifies to

Rmax=u2gcotθ.(2)(2)Rmax=u2gcot⁡θ.

Solving (1)(1) and (2)(2) will yield the desired expressions for θθ and RmaxRmax.