Why projectiles launched at complimentary angles have the same range?

Answer 1

You need to prove this.

find the range of a projectile of launch velocity V at Theta.
a. range=VcosTheta*timeinair
but hf=hi+vsinTheta*t-1/2 g t^2
or t(gt/2-vsinTheta)=0
timeinair= 2vsinTheta /t
range then= 2'V/g * cosTheta*sinTheta

range=V/g * sin(2*theta)

Now consider a new angle, PSI, PSI=90-theta

for the new angle,
range=V/g * sin(PSI)
range=V/g * sin (2*(90-theta))
= V/g * sin (180-2Theta)

Remember: Sin(a-b)= cosAsin(-b)+cos(-b)sinA
so sin (180-2theta)=cos180sin(-2theta)+sin180cos(-2theta)= -sin(-2theta)=sin2Theta

so, range for PSI and Theta are the same.

Answer 2

If you wish to remember a general equation for projectile motion, we can write: y=xtan(θ) –gx^2/[2(ucos(θ))^2] where u = initial velocity, θ = launch angle and the origin is the point of launch. Putting y=0 and solve for x to get: x=(u^2)sin(2θ)/g which well be written as x=(u^2)sin(2(90-θ))/g and, therefore, (θ)& (90-θ)produce the same range.