Pre calculus
posted by H .
An observer 2 km from the launching pad observes a vertically ascending missile at an angle of elevation of 21 degrees. 5 seconds later the nagle has increased to 35 degrees. What was its average speed during this interval? If it keeps going vertically at the same average speed, what will its angle of elevation be 15 seconds after the first sighting?

height at beginning of observation:
tan 21° = h/2 > height = 2tan21 = .76773
height after 5 seconds:
tan 35° = h/2 > height = 1.40042
change in height = .63269
average vertical speed = .63269/5 = .12654 km/s
so in 15 seconds it will have risen 1.89061 km on top of the .76773 km , its original height
so new height = 2.66579
tanØ = 2.66278/2 = 1.3328955..
Ø = 53.12°