Question

под каким углом к горизонту брошено тело если дальность в 2 раз больше Максимальной высоты подъёма? сопротивлением воздуха пренебречь СРОЧНО ПОЖАЛУЙСТА​

Answer 1

Надо использовать формулы дальности L и высоты h полёта.

L = 2Vo²sin(α)cos(α)/g.

h = Vo²sin(α)²/(2g).

На основе заданного условия: "дальность в 2 раз больше максимальной высоты подъёма" составим уравнение.

L = 2h, или, подставив значения, имеем:

2Vo²sin(α)cos(α)/g = 2Vo²sin(α)²/(2g).

После сокращения получаем 2cos(α) = sin(α).

Делим обе части на косинус угла.

2cos(α)/cos(α) = sin(α)/cos(α). Получаем:

2 = tg(α).

Отсюда α = arctg(2) ≈ 63,43495 градуса.

Answer 2

Angle of Projection of a Body Thrown at Twice the Maximum Height

To determine the angle at which a body is thrown with respect to the horizon if the range is twice the maximum height, we can use the principles of projectile motion.

In projectile motion, the range and maximum height of a projectile depend on the initial velocity and the angle of projection. However, in this case, we are given that the range is twice the maximum height and we need to find the angle of projection.

Let's assume the angle of projection is θ and the initial velocity is v.

The range of a projectile is given by the formula:

R = (v^2 * sin(2θ)) / g where R is the range, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

The maximum height of a projectile is given by the formula:

H = (v^2 * sin^2(θ)) / (2g) where H is the maximum height.

Given that the range is twice the maximum height, we can write:

2H = R

Substituting the formulas for R and H into the equation, we get:

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

Simplifying the equation, we have:

sin^2(θ) = sin(2θ) / 2

To solve this equation, we can use trigonometric identities. One such identity is:

sin(2θ) = 2sin(θ)cos(θ)

Substituting this identity into the equation, we get:

sin^2(θ) = (2sin(θ)cos(θ)) / 2

Simplifying further, we have:

sin^2(θ) = sin(θ)cos(θ)

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

sin(θ) = cos(θ)

Since sin(θ) = cos(θ), we can conclude that θ = 45 degrees.

Therefore, the body is thrown at an angle of 45 degrees with respect to the horizon if the range is twice the maximum height.

Please note that this calculation assumes no air resistance and uses basic principles of projectile motion.