Question

A man is moving away from a tower 41.6 m high at a rate of 2 m/s. If the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower, is

• A. $-\frac{4}{125}$ rad/s
• B. $-\frac{2}{25}$ rad/s
• C. $-\frac{1}{625}$ rad/s
• D. none of these

Answer 1

The correct option is A $-\frac{4}{125}$ rad/s

Since the angle of elevation is above the head of the man, hence
$tan\theta =\frac{41.6-1.6}{x}$
Or
$tan\theta =\frac{40}{x}$.
Or
$x=40cot\theta$
Differentiating with respect to time, we get
$v_x=-40cose{c}^2\theta .\frac{d\theta }{dt}$
Or
$2=-40cose{c}^2\theta .\frac{d\theta }{dt}$
Now
$cot\theta =\frac{30}{40}=\frac{3}{4}$ .. at that instant.
Hence
$cose{c}^2\theta =1+co{t}^2\theta =\frac{25}{16}$.
Substituting, we get
$-\frac{2}{40cose{c}^2\theta }=\frac{d\theta }{dt}$
Or
$-\frac{16.2}{40.25}=\frac{d\theta }{dt}$
Or
$-\frac{4}{125}rads/sec=\frac{d\theta }{dt}$