A man is walking under an inclined mirror at a constant velocity Vms-1.along the X axis. If the mirror is inclined at an angle θ with the horizontal, then what is the velocity of the image?.
A person measures the height of a building by walking out a distance of 46m from its base and shinning a flashlight beam toward the top. When the beam is elevated at angle of 39° w/ respect to the horizontal the beamstrikes the top of the building.
a. if the flashlight is held at a height of 2m, find the height of the building.
To determine the velocity of the image formed by the inclined mirror, we can use the concept of velocity transformations.
The velocity of the image can be found by resolving the velocity of the man relative to the mirror along the two perpendicular directions: one parallel to the mirror's surface (X-axis) and the other perpendicular to the mirror's surface (Y-axis).
Let's assume that the velocity of the man relative to the mirror along the X-axis is Vx and along the Y-axis is Vy.
Since the man is moving at a constant velocity V along the X-axis with respect to the ground, Vx is equal to V.
Now, since the mirror is inclined at an angle θ with the horizontal, we can use trigonometry to relate the velocities Vx and Vy.
We can express Vy in terms of V and θ using the trigonometric relationship:
Vy = V * sin(θ)
This means that the velocity of the man along the Y-axis, or the vertical direction, is V * sin(θ).
Since the image is formed due to the reflection of light rays, the velocity of the image will be determined by the direction and magnitude of the reflected rays. The reflected rays will have the same angle of incidence as the incident rays.
Therefore, the velocity of the image can be given as V' = -V * sin(θ).
The negative sign indicates that the image is formed on the opposite side of the mirror compared to the actual object.
Hence, the velocity of the image will be equal to -V * sin(θ).