Question

The maximum height of a projectile is twice of its horizontal range. Find the angle of projection? 1) 30 degrees 2) 45 degrees 3) 60 degrees 4) 90 degrees

Answer 1

To solve the Bernoulli differential equation y' - x/6y = x^9y^3, we use the substitution u = y^-2 to transform it into a linear differential equation in u. After solving for u, we revert the substitution to find y in terms of x and apply the initial condition y(1) = 1.

Step-by-step explanation:

The equation y' - x/6y = x9y3 is a Bernoulli differential equation, which is non-linear if the exponent n is neither 0 nor 1. In this case, the Bernoulli equation has n = 3. To solve it, we use the substitution u = y1-n or u = y-2. After the substitution, the differential equation becomes linear with respect to u, and we solve for u before finding y.

To perform the substitution, we differentiate u with respect to x, which gives us du/dx = -2y-3y'. Substituting y = u-1/2 and y' = -1/2 u-3/2du/dx into the original equation, we obtain a linear differential equation in u. After simplifying and integrating, we find u as a function of x. Subsequently, we convert back to find y in terms of x and apply the initial condition y(1) = 1 to find the particular solution.

Answer 2

The angle of projection for the given scenario is 60 degrees, thus the correct option is 3.

Step-by-step explanation:

When a projectile is launched at an angle θ to the horizontal with initial velocity
`\( v_0 \)`, its maximum height
`(\( H_(max) \))` and horizontal range (\( R \)) can be calculated using projectile motion equations. According to the problem, the maximum height of the projectile is twice its horizontal range, which can be represented as
`\( H_(max)` = 2R.

The formulas for the maximum height and horizontal range in terms of initial velocity
`(\( v_0 \))` and angle of projection (θ) are:

`\( H_(max) = (v_0^2 \cdot \sin^2θ)/(2g) \)` (Formula for maximum height)

`\( R = (v_0^2 \cdot \sin(2θ))/(g) \)` (Formula for horizontal range)

From the given information, equating
`\( H_(max) \)` and 2R, we get:

`{v_0^2 \cdot \sin^2`θ/2g= 2 ·
`(v_0^2 \cdot \sin(2θ))/(g)`

Solving for θ, we find:

sin²θ = 4 c·sin(2θ)

sin²θ = 4 · 2 · sinθ c· cosθ

sinθ = 8· cosθ

By using trigonometric identities, sinθ = 8· cosθ simplifies to tanθ = 1/8. Solving for θ, we get θ =
`\arctan\left((1)/(8)\right)` ≈ 7.12
`5^\circ \)`. Since θ represents the angle of projection, which is greater than 45 degrees, the nearest option is 60 degrees, hence the correct angle of projection is 60 degrees (option 3).