Question

A vertical pole stands in an empty pond and its shadow is formed on the bottom, the sun being oº over the horizon. At t = 0 water inlet of the pond is opened and the height of water (u - g) in the pond increases at a constant rate of 1 cm/sec. As the level of the water increases, length of shadow formed on the bottom decreases till t = 300 sec. At t > 300 sec, length of shadow is same as at t = 300 sec. Further length of shadow at t = 300 see is the same as it would be if the sun is 60° over the horizon and pond is empty. Length of pole is​

Answer 1

Let OA be the vertical pole of height h and OP
1
​
,OP
2
​
and OP
3
​
be the lengths of shadow.
In Δ AOP
1
​
, we have
tanθ
1
​
=
OP
1
​

OA
​
=
h
h
​
=1⇒θ
1
​
=
4
π
​

In ΔAOP
2
​
, we have
tanθ
2
​
=
OP
2
​

OA
​
=
2h
h
​
=
2
1
​

θ
2
​
=tan
−1
1/2
Similarly, in ΔAOP
3
​
, we have
tanθ
3
​
=1/3
θ
3
​
=tan
−1
(1/3)
Therefore, sum of the angles of elevation of the eyes
=θ
1
​
+θ
2
​
+θ
3
​

=
4
π
​
+tan
−1
(
2
1
​
)+tan
−1
(
3
1
​
)
=
4
π
​
+tan
−1

⎝
⎜
⎜
⎛
​

1−
3
1
​
×
2
1
​

2
1
​
+
3
1
​

​

⎠
⎟
⎟
⎞
​
[∵tan
−1
x+tan
−1
y=tan
−1
(
1−xy
x+y
​
)]
=
4
π
​
+tan
−1
(
5/6
5/6
​
)
=
4
π
​
+tan
−1
(1)=
4
π
​
+
4
π
​
=
2
π