SOLUTION: A hot-air balloon is released at 1:00 P.M. and rises vertically at a rate of 4 meters/sec . An observation point is situated at 125 meters from a point on the ground directly belo

Algebra ->  Functions -> SOLUTION: A hot-air balloon is released at 1:00 P.M. and rises vertically at a rate of 4 meters/sec . An observation point is situated at 125 meters from a point on the ground directly belo      Log On


   

Question 127638This question is from textbook algebra and trigonometry with analiytic geometry
: A hot-air balloon is released at 1:00 P.M. and rises vertically at a rate of 4 meters/sec . An observation point is situated at 125 meters from a point on the ground directly below the balloon (see the figure). If t denotes the time (in seconds) after 1:00 P.M., express the distance d between the balloon and the observation point as a function of t.
Please use the notation "sqr(a)" which equals PLEASE PLEASE, HELP This question is from textbook algebra and trigonometry with analiytic geometry

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The observation point, the point directly below and the
height of the ballon form a right triangle. The distance
from the observer to the ballon is the hypotenuse
d+=+sqrt%28h%5E2+%2B+125%5E2%29where distances are in meters
The ballon is going up at a rate of 4 m/sec. If t
is the elapsed time, h+=+4twhere tis in seconds
and h is in meters.
d+=+sqrt%28%284t%29%5E2+%2B+125%5E2%29
d+=+sqrt%2816t%5E2+%2B+15625%29 meters answer