The projectile motion is fired with velocity of magnitude vo at the angle feta. Find feta for which the maximum elevation of the projectile is twice its range

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To find the angle θ at which the maximum elevation of the projectile is twice its range, we can use the equations for projectile motion.

Let's start by defining the variables:
- vo: initial velocity magnitude
- θ: launch angle
- g: acceleration due to gravity (assumed to be 9.8 m/s²)
- R: range of the projectile
- H: maximum elevation of the projectile

The range of the projectile is given by the formula:
R = (vo² * sin(2θ)) / g

The maximum elevation of the projectile is given by the formula:
H = (vo² * sin²(θ)) / (2g)

We are given that H = 2R, so we can substitute these equations into each other:

(vo² * sin²(θ)) / (2g) = 2 * (vo² * sin(2θ)) / g

Now we can simplify the equation to solve for θ:

sin²(θ) = 4 * sin(2θ)

Using the trigonometric identity sin(2θ) = 2 * sin(θ) * cos(θ), we can rewrite the equation as:

sin²(θ) = 8 * sin(θ) * cos(θ)

Divide both sides by sin(θ) and cancel out common factors:

sin(θ) = 8 * cos(θ)

Now, rearrange the equation:

sin(θ) / cos(θ) = 8

tan(θ) = 8

Taking the inverse tangent of both sides, we get:

θ = tan^(-1)(8)

Using a scientific calculator, the value of θ is approximately 83.54 degrees (rounded to two decimal places).

Therefore, the angle θ at which the maximum elevation of the projectile is twice its range is approximately 83.54 degrees.