Why projectiles launched at complimentary angles have the same range?

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.

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.

When projectiles are launched at complimentary angles (angles that add up to 90 degrees, such as 30 degrees and 60 degrees), they will have the same range because the vertical components of their velocities cancel each other out.

To understand why this happens, let's break down the motion of a projectile. The motion can be divided into two components: horizontal and vertical. The horizontal motion remains constant and unaffected by gravity, while the vertical motion is affected by gravity.

When a projectile is launched at an angle, it has an initial velocity that can be broken down into two components: the horizontal component (Vx) and the vertical component (Vy).

Now, when a projectile is launched at a complimentary angle, let's say one at 30 degrees (angle A) and the other at 60 degrees (angle B), the vertical components of their velocities would be Vsin(A) and Vsin(B), respectively, where V is the initial velocity of the projectile.

Since the sine of an angle is equal to the sine of its complimentary angle (sin(A) = sin(90-A)), we can say that sin(A) = sin(B). Therefore, the vertical components of the velocities of the two projectiles are equal.

Now, as the projectiles follow their trajectories, the gravitational force affects their vertical motion. Gravity acts downward, causing a downward acceleration (g) in both projectiles. However, since the vertical velocities of the projectiles are equal, the effect of gravity on their vertical motions is the same.

As a result, both projectiles would reach the same maximum height and take the same amount of time to reach the ground. Since the horizontal component of their velocities remains constant, the range they cover (the horizontal distance traveled) would also be the same.

In summary, projectiles launched at complimentary angles have the same range because the cancellation of their vertical velocities results in the same effect from gravity, leading to the same maximum height and time of flight. The horizontal velocity remains constant, which means the projectiles cover the same horizontal distance or range.