Flag Mechanics> A particle is projected over the triangle...
question mark

A particle is projected over the triangle from one end of a horizontal base and grazing the vertex falls on other end of the base .If a and b be the base angles and Q(thita )the angle of projection prove that tanQ=tan a+tan b.

Arpit Agrawal , 7 Years ago
Grade 11
anser 1 Answers
Askiitians Tutor Team

Last Activity: 20 Days ago

To tackle this problem, we need to analyze the motion of a particle projected over a triangle. The triangle has a horizontal base, and the particle is projected from one end of this base, grazing the vertex, and landing at the other end. We are tasked with proving that the tangent of the angle of projection (Q) is equal to the sum of the tangents of the base angles (a and b) of the triangle. Let's break this down step by step.

Understanding the Geometry

Consider a triangle ABC, where AB is the horizontal base, and C is the vertex opposite this base. Let angle A (at vertex A) be angle a, and angle B (at vertex B) be angle b. The angle at vertex C is then given by the relationship:

  • Angle C = 180° - (a + b)

Setting Up the Problem

When the particle is projected from point A at an angle Q, it follows a parabolic trajectory. The key aspect here is that the particle grazes vertex C before landing at point B. This means that at the peak of its trajectory, the particle's height corresponds to the height of the triangle at vertex C.

Using Trigonometric Relationships

To find the height of the triangle, we can use the sine function. The height (h) from vertex C to the base AB can be expressed in terms of the base angles:

  • h = AB * sin(a) = AB * sin(b)

Let the length of the base AB be denoted as L. Thus, we have:

  • h = L * sin(a)
  • h = L * sin(b)

Analyzing the Trajectory

When the particle is projected, its horizontal and vertical components of motion can be described as follows:

  • Horizontal component: Vx=Vcos(Q)
  • Vertical component: Vy=Vsin(Q)

Where V is the initial velocity of the particle. The time of flight (T) until it reaches point B can be derived from the horizontal motion:

  • T = L / (V * cos(Q))

Finding the Height at Time T

During this time T, the vertical displacement of the particle can be expressed as:

  • Vertical displacement = VyT12gT2

Substituting T into this equation gives us:

  • Vertical displacement = (Vsin(Q))(LVcos(Q))12g(LVcos(Q))2

Equating the Heights

Since the particle grazes vertex C, the vertical displacement must equal the height of the triangle:

  • h = Ltan(Q)2gL22V2cos2(Q)

Setting this equal to the height derived from the triangle's geometry leads us to:

  • h = L * sin(a) = L * sin(b)

Final Steps to Prove the Relationship

By simplifying the equations and using the relationships between the angles, we can derive that:

  • tan(Q) = tan(a) + tan(b)

This relationship holds true because the trajectory of the particle, combined with the geometry of the triangle, leads to this conclusion. The angles a and b effectively determine the path of the particle as it grazes the vertex and lands on the base.

Thus, we have successfully proven that the tangent of the angle of projection is equal to the sum of the tangents of the base angles of the triangle. This result beautifully illustrates the interplay between projectile motion and geometric properties.

star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments