# Thread: angular rate of change

1. ## angular rate of change

in a free fall experiment, an object is dropped from a height of 256 ft. a camera on the ground 500 ft from the point of impact records the fall of the object. find the rates of change of the angle of elevation of the camera when t = 1 and t = 2. im not sure how to start this problem. i found the position function which was pretty easy but im not sure how to start this one. s(t) = -16t^2 + 256

2. Originally Posted by slapmaxwell1
in a free fall experiment, an object is dropped from a height of 256 ft. a camera on the ground 500 ft from the point of impact records the fall of the object. find the rates of change of the angle of elevation of the camera when t = 1 and t = 2. im not sure how to start this problem. i found the position function which was pretty easy but im not sure how to start this one. s(t) = -16t^2 + 256
let $\displaystyle \theta$ be the camera's angle of elevation.

let $\displaystyle h$ = object's height above the ground

set up a trig ratio with a right triangle ...

$\displaystyle \tan{\theta} = \frac{h}{500}$

take the time derivative ...

$\displaystyle \sec^2{\theta} \cdot \frac{d\theta}{dt} = \frac{1}{500} \cdot \frac{dh}{dt}$

$\displaystyle \frac{d\theta}{dt} = \cos^2{\theta} \cdot \frac{1}{500} \cdot \frac{dh}{dt}$

use the equation

$\displaystyle h = 256-16t^2$

to find the required values needed in the $\displaystyle \frac{d\theta}{dt}$ equation, then evaluate $\displaystyle \frac{d\theta}{dt}$ at the given times.