Trace the path of a ray emitted from the tip of the object through the focal point of the mirror and then the reflected ray that
results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray. Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded.
1 answer:
Answer:
1 / f = 1 / p + 1 / q
Explanation:
Mirror problems can be solved analytically with the constructor equation
1 / f = 1 / p + 1 / q
An approximate solution can also be given using graphical methods, where two of three rays must be drawn,
1) A ray that passes through the focal length and bounces horizontally
2) A ray that passes through the center of curvature (c = 2f) of the mirror and does not deviate
3) A beam that comes horizontal and bounces through the focal
The attached shows the ray that passes through the focal length and bounces horizontally.
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Answer:
Approximately (assuming that the projectile was launched at angle of
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing is the same as the altitude
at which this projectile was launched:
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is (upwards,) the vertical velocity right before landing would be
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be . In other words,
.
Hence, the time it takes to achieve a (vertical) velocity change of would be:
.
Hence, this projectile would be in the air for approximately .