A projectile is fired in such a way that its horizontal range is equal to 2.6 times its maximum height, what is the angle of projection?
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To find the angle of projection, we need to analyze the horizontal range and maximum height of the projectile.
Let's assume the initial velocity of the projectile is "v" and the angle of projection is "θ".
The horizontal range (R) of a projectile can be calculated using the formula:
R = (v^2 * sin(2θ)) / g
where g is the acceleration due to gravity.
The maximum height (H) of a projectile can be calculated using the formula:
H = (v^2 * sin^2(θ)) / (2g)
Given that the horizontal range is 2.6 times the maximum height, we can write the equation as follows:
2.6H = R
Substituting the formulas for R and H, we have:
2.6 * ((v^2 * sin^2(θ)) / (2g)) = ((v^2 * sin(2θ)) / g)
Cancelling out v^2 and g from both sides, we have:
2.6 * sin^2(θ) = sin(2θ)
Using the trigonometric identity sin(2θ) = 2 * sin(θ) * cos(θ), we can rewrite the equation as:
2.6 * sin^2(θ) = 2 * sin(θ) * cos(θ)
Divide both sides of the equation by 2 and sin(θ), we get:
1.3 * sin(θ) = cos(θ)
Divide both sides of the equation by cos(θ), we have:
tan(θ) = 1.3
Using the inverse tangent function (tan^(-1)), we can find the angle of projection:
θ = tan^(-1)(1.3)
Using a calculator, we find:
θ ≈ 52.43 degrees
Therefore, the angle of projection is approximately 52.43 degrees.