A stone projected at angle alpha with plane
inclined at angle theta find relation between
alpha and theta if stone hit at right angke with
the inclined plane?
To find the relation between the angle of projection (α) and the angle of inclination of the plane (θ), so that the stone hits the plane at a right angle, we can use the concept of the relative angle of incidence and reflection.
Let's assume that the stone is projected from a horizontal plane with an angle of projection α. The plane is inclined at an angle θ. When the stone hits the plane at a right angle, the angle of incidence (i) will be equal to the angle of reflection (r) with respect to the surface of the plane.
The angle of incidence is the angle between the direction of motion of the stone and the normal to the plane at the point of impact. Similarly, the angle of reflection is the angle between the direction of motion of the stone after hitting the plane and the normal to the plane at the point of impact.
Since the stone hits the plane at a right angle, the angle of incidence (i) will be 90 degrees. Therefore, to find the relation between α and θ, we need to determine the value of α when the angle of incidence is 90 degrees.
To calculate the angle of incidence (i), we need to use the concept of the relative angle between the stone's initial velocity and the normal to the plane. The relative angle between two vectors A and B is given by:
cos(theta) = (A · B) / (|A| |B|)
In this case, vector A represents the stone's initial velocity, and vector B represents the normal to the inclined plane.
The direction of the stone's initial velocity can be represented as a vector with components (u cos α) in the horizontal direction and (u sin α) in the vertical direction, assuming the initial speed of the stone is u.
Now, let's consider the normal to the inclined plane. This vector will depend on the angle of inclination θ. Assuming the normal points away from the inclined plane, its components can be represented as (sin θ) in the horizontal direction and (-cos θ) in the vertical direction.
Calculating the dot product (A · B) of the stone's initial velocity and the normal to the plane:
(A · B) = (u cos α)(sin θ) + (u sin α)(-cos θ)
To calculate the magnitudes of vectors A and B:
|A| = √((u cos α)^2 + (u sin α)^2) = √(u^2 cos^2 α + u^2 sin^2 α) = u
|B| = √((sin θ)^2 + (-cos θ)^2) = √(sin^2 θ + cos^2 θ) = 1
Substituting these values into the equation for the relative angle:
cos(i) = [(u cos α)(sin θ) + (u sin α)(-cos θ)] / (u * 1)
Since we know that i = 90 degrees:
cos(90) = [(u cos α)(sin θ) + (u sin α)(-cos θ)] / u
0 = [(u cos α)(sin θ) + (u sin α)(-cos θ)] / u
0 = cos α sin θ - sin α cos θ
Using the trigonometric identity sin(x - y) = sin x cos y - cos x sin y:
0 = sin(θ - α)
To satisfy this equation, the angle (θ - α) must be equal to 0 or any multiple of π radians (180 degrees). In other words, θ - α = 0 + nπ, where n is an integer.
Therefore, the relation between α and θ, when the stone hits the inclined plane at a right angle, is:
θ = α + nπ