find the angle of projection for which max height of a projectile is equal to half of its horizontal range?

This article should get you off the ground, as it were ...

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To find the angle of projection for which the maximum height of a projectile is equal to half of its horizontal range, we can use the equations of motion for projectile motion.

Let's assume that the initial velocity of the projectile is 'v' and the angle of projection is 'θ'. The horizontal range of the projectile, R, is given by the formula:

R = (v^2 * sin(2θ)) / g

where g is the acceleration due to gravity.

The maximum height, H, can be calculated using the formula:

H = (v^2 * sin^2(θ)) / (2g)

According to the problem, we need to find the angle of projection for which H is equal to half of R. Therefore, we can set up the following equation:

H = (R / 2)

Substituting the values of H and R from the above equations, we have:

(v^2 * sin^2(θ)) / (2g) = (v^2 * sin(2θ)) / (2g)

By simplifying and canceling out the common terms, we get:

sin^2(θ) = sin(2θ) / 2

Using the double-angle formula for sine, sin(2θ) = 2sin(θ)cos(θ), we can rewrite the equation as:

sin^2(θ) = (2sin(θ)cos(θ)) / 2

sin^2(θ) = sin(θ)cos(θ)

Dividing both sides by sin(θ), we get:

sin(θ) = cos(θ)

Now, we can solve this equation to find the angle of projection θ. Taking the inverse sine of both sides, we have:

θ = arcsin(cos(θ))

To find the value of θ, we can use numerical methods or trial and error. One common approach is to graph the two functions on a coordinate plane - y = arcsin(cos(x)) and y = x - and find the point of intersection.

Alternatively, using a scientific calculator or mathematical software, we can use an iterative numerical method to find the value of θ. One such method is the Newton-Raphson method or the bisection method.

By applying these methods, we can determine the angle of projection for which the maximum height of the projectile is equal to half of its horizontal range.