A balloon rises vertically from a point on level ground at the rate of 3 m/s. An observer stands on the ground 40 m away from the point of release. The rate at which the angle of elevation of the balloon, as seen by the observer, is increasing when the balloon is 30 m high is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{6}{125}\) radian/s
Let h be the height of the balloon and θ the angle of elevation at time t. The horizontal distance stays 40 m, so
\( \displaystyle \tan\theta = \frac{h}{40} \)
Differentiating with respect to t (the rates of change are linked):
\( \displaystyle \sec^{2}\theta \, \frac{d\theta}{dt} = \frac{1}{40} \cdot \frac{dh}{dt} \)
When h = 30:
\( \displaystyle \tan\theta = \frac{30}{40} = \frac{3}{4} \)
\( \displaystyle \sec^{2}\theta = 1 + \frac{9}{16} = \frac{25}{16} \)
With \(\frac{dh}{dt} = 3\):
\( \displaystyle \frac{d\theta}{dt} = \frac{3}{40} \times \frac{16}{25} \)
\( \displaystyle = \frac{48}{1000} = \frac{6}{125} \) radian/s
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3}{40}\) radian/s — Incorrect. This is \(\frac{1}{40} \cdot \frac{dh}{dt}\) alone; it leaves out the factor \(\sec^{2}\theta = \frac{25}{16}\).
Option (2) \(\displaystyle\frac{9}{250}\) radian/s — Incorrect. This is \(\frac{30 \times 3}{2500}\): it uses the height 30 m in place of the distance 40 m.
Option (3) \(\displaystyle\frac{3}{50}\) radian/s — Incorrect. This is \(\frac{3}{40} \times \frac{4}{5}\): it multiplies by \(\cos\theta\) instead of \(\cos^{2}\theta = \frac{16}{25}\).