How Does Projectile Angle Affect Range in Physics?

[Questions999]

Galileo showed that, if air resistance is neglected, the ranges for projectiles whose angles of projection exceed or
fall short of 45 degree by the same amount are equal. Prove Galileo’s result.
Answer in my textbook : set θ=45+ - ∂ (I understand this)
Now comes the weird part R = [(v0^2)*sin(90-2∂ ]/g=[V0^2*cos(+-2∂ )]/g.. What formula is this,and how is it transformed this way?

since cos2∂ =cos(-2∂ )
Now we have R(45+θ)=R(45-θ) ..how about this one?How did we get these?

 

[voko]

You need the formula for range given the magnitude of initial velocity and its angle with the horizontal. Do you have it? If not, you will have to derive it from other equations that you have.

 

[Questions999]

I have them but I don't know how to relate them :(

 

[voko]

What equations do you have?

 

[Nickie]

The formula used here is the range equation for projectile motion, which is derived from the equations of motion and the kinematic equations. It states that the horizontal distance traveled by a projectile (R) is equal to the initial velocity (v0) squared, multiplied by the sine of twice the launch angle (2θ), divided by the acceleration due to gravity (g).

In the textbook, the angle of projection (θ) is set to be 45 degrees plus or minus a small change (∂). This means that the launch angle can be written as 45+∂ for one case and 45-∂ for the other case.

Using the trigonometric identity cos2∂ = cos(-2∂), we can rewrite the formula as R = (v0^2 * cos(2θ)) / g. Since we have set θ to be 45+∂ or 45-∂, we can substitute these values in the formula to get R(45+θ) and R(45-θ).

This shows that the ranges for angles exceeding or falling short of 45 degrees by the same amount (∂) are equal, proving Galileo's result. This is because the launch angle is the only variable that affects the range in this formula, and by setting it to be 45+∂ or 45-∂, we are essentially launching the projectile at the same angle but in different directions (either slightly above or slightly below 45 degrees). Since the formula shows that the range is the same for both cases, it proves Galileo's result that the ranges are equal for angles exceeding or falling short of 45 degrees by the same amount.