from a point 1.5 mi from a launcch pad at cape canaveral, an observer sights a space shuttle at an angle of elevation of 10 degrees moments after it is launched. after 10 seconds, the angle of elevation is 50 degrees. how far does the space shuttle travel vertically during the 10 second interval? what is the average speed of the space shuttle during that interval?

Given angles of elevation of 10° and 50° measured 10 seconds apart from an observation point 1.5 miles away, you want to know the vertical change in distance, and the average speed of the shuttle between the two observations.

Height

The tangent relation tells you ...

Tan = Opposite/Adjacent

Then the height of the shuttle at the observed angles was ...

Opposite = Tan · Adjacent

h1 = tan(10°)·(1.5 mi) ≈ 0.264490 mi

h2 = tan(50°)·(1.5 mi) ≈ 1.787630 mi

Vertical travel

The amount of vertical travel in the 10-second interval was ...

1.787630 mi -0.264490 mi = 1.523240 mi

The space shuttle traveled about 1.523 miles vertically in the 10-second interval.

Speed

In miles per second, the average speed during that time was about ...

(1.523 mi)/(10 s) = 0.1523 mi/s

In miles per hour, that's about ...

(0.1523 mi/s)(3600 s/h) ≈ 548 mi/h