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Show that the range of a projectile for two angles of projection alpha and beta is the same, where alpha + beta is equal to 90 degrees.

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Final answer:

The range of a projectile is equal for two angles of projection alpha and beta that sum to 90 degrees, due to the trigonometric identity sin(90° - θ) = cos(θ), which leads to equal range expressions when substituting beta (90° - alpha) into the range equation.

Step-by-step explanation:

Show the Range for Two Complementary Angles

When analyzing the range of a projectile for two angles of projection α (alpha) and β (beta) where α + β = 90°, we'll need to use the standard range equation for projectile motion without air resistance:


R = ​​(​(vo²) * sin(2θ)/g), where 'R' is the range, 'vo' is the initial velocity, 'θ' is the initial angle, and 'g' is the acceleration due to gravity (approximately 9.81 m/s²).

If we take two angles α and β that are complementary (adding up to 90°), we can use the identity sin(90° - θ) = cos(θ). Hence, for a given α, β will be 90° - α.

Now, let's calculate the range for α:

Rα = (vo²) * sin(2α)/g

For β:

Rβ = (vo²) * sin(2β)/g
= (vo²) * sin(2(90° - α))/g
= (vo²) * sin(180° - 2α)/g

Since sin(180° - θ) is equal to sin(θ), we have:

Rβ = (vo²) * sin(2α)/g

Thus, Rα = Rβ, meaning the ranges are equal for two complementary angles.

User Haseeb Saeed
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