A cannon that is capable of firing a shell at speed V0 is mounted on a vertical

Question

A cannon that is capable of firing a shell at speed V0 is mounted on a vertical tower of height
h that overlooks a level plain below.
(a) Show that the elevation angle a at which the cannon must be set to achieve maximum
range is given by the expression

(b) What is the maximum range R of the cannon?

I got part (a) pretty easily but part b is my problem.

I got the value of R to be:

The answer is suppoed to be:

I know I am supposed to eliminate the angle between the two equations, but I can't figure out how to do it.

Thanks in advance!

Explanation / Answer

speed of cannon ball shot from cannon = Vo

height of vertical tower = h

a. let angle of elevation be theta

then vertical speed = vsin(theta)

time taken to reach max height, t1 , maximum height h'

2*h'*g = v^2sin^2(theta)

vsin(theta) = gt1

t1 = vsin(theta)/g

h' = v^2*sin^2(theta)/2g

tinme taken to decend form this height = t2

h + h' = 0.5*g*t2^2

t2 = sqroot(2(h + v^2*sin^2(theta)/2g)/g)

total time of flight = t1 + t2 = vsin(theta)/g + sqroot(2(h + v^2*sin^2(theta)/2g)/g)

range = R = [vsin(theta)/g + sqroot(2(h + v^2*sin^2(theta)/2g)/g)]*vcos(theta)

now dR/d(theta) = 0 for maximum range

-[vsin(theta)/g + sqroot(2(h + v^2*sin^2(theta)/2g)/g)]*vsin(theta) + [vcos(theta)/g + ((v^2*sin(theta)cos(theta))/g^2*sqroot(2(h + v^2*sin^2(theta)/2g)/g)]*vcos(theta) = 0

vsin^2(theta)/g + sin(theta)*sqroot(2(h + v^2*sin^2(theta)/2g)/g) = vcos^2(theta)/g + cos(theta)((v^2*sin(theta)cos(theta))/g^2*sqroot(2(h + v^2*sin^2(theta)/2g)/g)]

solving this we get

sin^2(alpha) = vo^2/(2(vo^2 + gh))

or

csc^2(alpha) = 2(1 + gh/vo^2) [ where alpha is theta for maximum range]

b. max range = R

R = [v^2/g*sqrt(2(v^2 + gh)) + sqroot(2(h + v^4/4g(v^2 + gh))/g)]*vsqroot(1 - v^2/2(v^2 + gh))

this can be simlified to get the desired answer

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