Math Problem Statement

Камень брошен горизонтально со склона горы, образующего угол α = 45° с горизонтом. Чему равна начальная скорость v₀ камня (м/с), если он упал на склон на расстоянии L = 50 м от точки бросания?

Solution

To solve this problem, we need to analyze the motion of the stone as it is projected horizontally from the slope.

Key Data:

  1. The slope angle with the horizontal, α=45\alpha = 45^\circ.
  2. The stone lands on the slope at a distance L=50mL = 50 \, \text{m} from the projection point.
  3. The stone is projected horizontally with initial speed v0v_0, and the task is to determine v0v_0.

Analysis:

We approach this using the equations of motion and the geometry of the situation.

  1. Coordinate System:

    • Choose a coordinate system where:
      • xx' is parallel to the slope.
      • yy' is perpendicular to the slope.
    • Transform the motion into this coordinate system.
  2. Equations of Motion: In the regular horizontal-vertical coordinate system (xx-yy):

    • Horizontal motion: x=v0tx = v_0 t,
    • Vertical motion: y=12gt2y = \frac{1}{2} g t^2, where gg is the acceleration due to gravity (9.8m/s29.8 \, \text{m/s}^2).
  3. Slope Geometry: The equation of the slope in the horizontal-vertical plane: y=xtanαy = x \cdot \tan\alpha Using α=45\alpha = 45^\circ, this simplifies to: y=xy = x

    The condition for the stone to land on the slope is: 12gt2=v0t\frac{1}{2} g t^2 = v_0 t At the same time, x2+y2=L2x^2 + y^2 = L^2, or equivalently, x=yx = y and: 2x2=L22x^2 = L^2


Solving for v0v_0:

  1. From the geometry of the slope: x=y=L2x = y = \frac{L}{\sqrt{2}}

  2. Time of flight tt: From horizontal motion, t=xv0t = \frac{x}{v_0}, and from vertical motion, y=12gt2y = \frac{1}{2} g t^2. Equating these: L2=12g(L2v0)2\frac{L}{\sqrt{2}} = \frac{1}{2} g \left(\frac{L}{\sqrt{2} v_0}\right)^2 Simplify: v0=gL2v_0 = \sqrt{\frac{gL}{\sqrt{2}}}


Final Calculation:

Substitute g=9.8m/s2g = 9.8 \, \text{m/s}^2 and L=50mL = 50 \, \text{m}: v0=9.8502v_0 = \sqrt{\frac{9.8 \cdot 50}{\sqrt{2}}} v0=4901.414346.4718.61m/s.v_0 = \sqrt{\frac{490}{1.414}} \approx \sqrt{346.47} \approx 18.61 \, \text{m/s}.


Result:

The initial speed of the stone is: v018.6m/s.v_0 \approx 18.6 \, \text{m/s}.

Would you like detailed steps or further exploration of this solution?

Here are 5 related questions:

  1. How does the solution change if the slope angle is different?
  2. What would happen if air resistance is considered in this scenario?
  3. How can we calculate the flight time of the stone in this motion?
  4. How does the initial velocity vary if the landing distance LL is doubled?
  5. Can we solve the problem using energy conservation principles?

Tip: Always double-check geometry and trigonometric relationships when dealing with inclined motion!

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Math Problem Analysis

Mathematical Concepts

Projectile motion
Inclined planes
Kinematics in physics

Formulas

x = v₀ * t
y = 0.5 * g * t²
y = x * tan(α)
x² + y² = L²
v₀ = sqrt((g * L) / sqrt(2))

Theorems

Equations of motion for uniform acceleration
Trigonometric relationships in inclined planes

Suitable Grade Level

Grades 10-12 or introductory university physics