Question

A rocket reaches a height of 300\/3 m. A crowd observing the launch stands 300 m away. What is the maximum angle, e, in which the crowd observes the launch?
a)
60°
b)
30°
C) 75°
d) 45°

Answer 1

To determine the maximum angle \( e \) at which the crowd observes the launch, we can use the tangent function in a right triangle formed by the height of the rocket and the distance from the crowd to the base of the rocket.

Given:
- Height \( h = \frac{300}{3} = 100 \) m
- Distance \( d = 300 \) m

The tangent of angle \( e \) is given by:

\[
\tan(e) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d} = \frac{100}{300} = \frac{1}{3}
\]

To find \( e \), we need to compute the arctangent (inverse tangent) of \( \frac{1}{3} \):

\[
e = \tan^{-1}\left(\frac{1}{3}\right)
\]

Using a calculator or trigonometric tables, we can find:

\[
e \approx 18.43^\circ
\]

None of the provided options match this value directly. However, assuming the options are approximate or simplified, it would be reasonable to choose the closest angle from the list.

Among the options provided:
- a) 60°
- b) 30°
- c) 75°
- d) 45°

The closest value to 18.43° is actually not listed. Therefore, none are correct but if forced to choose based on the options available, none realistically apply.

Please double-check the question or the available answer choices for any discrepancies.