A plane mirror of circular shape with radius r=20cm is fixed to the ceiling. A bulb is to be placed on the axis of the mirror. A
circular area of r=1m on the floor is to be illuminated after reflection of light from mirror.The height of the room is 3m.What is maximum distance from the centre of the mirror and the bulb so that the required area is illuminated?
1 answer:
Answer:
0.75 m
Explanation:
Let's call the distance between the bulb and the mirror x.
The bulb and the length of the mirror form a triangle. The mirror and the illuminated area on the floor form a trapezoid. If we extend the lines from the mirror edge to the reflected image of the bulb, we turn that trapezoid into a large triangle. This triangle and the small triangle are similar. So we can say:
x / 0.4 = (3 + x) / 2
Solving for x:
2x = 0.4 (3 + x)
2x = 1.2 + 0.4 x
1.6 x = 1.2
x = 0.75
So the bulb should located no more than 0.75 m from the mirror.
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Answer:
∅ = 89.44°
Explanation:
In situations like this air resistance are usually been neglected thereby making g= 9.81 m/
Bring out the given parameters from the question:
Initial Velocity () = 1000 m/s
Target distance (d) = 2000 m
Target height (h) = 800 m
Projection angle ∅ = ?
Horizontal distance = tcos ∅ .......................... Equation 1
where = velocity in the X - direction
t = Time taken
Vertical Distance = y = t -
g
................... Equation 2
Where = Velocity in the Y- direction
t = Time taken
=
sin∅
Making time (t) subject of the formula in Equation 1
t = d/(cos ∅)
t = =
=
...................Equation 3
substituting equation 3 into equation 2
Vertical Distance = d =
-
g
Vertical Distance = h = sin∅ -
g
Vertical Distance = h = dtan∅ - g
Applying geometry
=
+ 1
Vertical Distance = h = d tan∅ - 2 g ( + 1)
substituting the given parameters
800 = 2000 tan ∅ - 2 (9.81)( + 1)
800 = 2000 tan ∅ - 19.6( + 1) Equation 4
Replacing tan ∅ = Q .....................Equation 5
In order to get a quadratic equation that can be easily solve.
800 = 2000 Q - 19.6 + 19.6
Rearranging 19.6 - 2000 Q + 780.4 = 0
= 101.6291
= 0.411
Inserting the value of Q Into Equation 5
tan ∅ = 101.63 or tan ∅ = 0.4114
Taking the Tan inverse of each value of Q
∅ = 89.44° ∅ = 22.37°