Prove that for angle of projection, which exceed or fall short of 45° by equal amount the ranges are equal.



Given:

Angle = θ =  45°


To Prove:

Exceed or fall short of 45° by equal amount the ranges are equal



Solution:

Exceed of 45°

Let θ = 45° + Ñ„, The Range R₁

R₁ = `\frac {vi² sin (2θ₁)}{g}`

R₁ = `\frac {vi² sin (2(45° + Ñ„))}{g}`

R₁ = `\frac {vi² sin (90° + 2Ñ„)}{g}`

R₁ = `\frac {vi² sin (90° + 2Ñ„)}{g}`

But sin (90° + 2Ñ„) = cos 2Ñ„ hence

R₁ = `\frac {vi² cos 2Ñ„}{g}` ------------(1)

Fall short of 45°

Let θ = 45° - Ñ„, The Range R₂

R₂ = `\frac {vi² sin (2θ₂)}{g}`

R₂ = `\frac {vi² sin (2(45° - Ñ„))}{g}`

R₂ = `\frac {vi² sin (90° - 2Ñ„)}{g}`

R₂ = `\frac {vi² sin (90° - 2Ñ„)}{g}`

But sin (90° - 2Ñ„) = cos 2Ñ„ hence

R₂ = `\frac {vi² cos 2Ñ„}{g}` ------------(2)

By comparing equation (1) an (2)

R₁ = R₂

Hence the required proof


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