# Angle of line of sight for a vertically lauched rocket

1. Jan 8, 2005

### ziddy83

Hey...whats up,
ok... here's the problem.

A rocket is launched vertically from a point on the ground that is 100 horizontal meters from an observer with binoculars. The rocket is rising vertically and its height above the ground (in meters) is given by : $$y(t)=60t-5t^2$$
Two seconds after launch, how fast must the observer change the angle of elevation of her line of sight to keep the rocket in the binoculars?

I drew out the figure and...i know that i need to somehow compare the rate of change of the height of the rocket and the angle of her line of sight. So the rate of change of the height is y ' , which i got $$y' = t - 10t$$

now how can i relate the two? I can plug in 2 seconds in y' to get the rate at wich the height is changing, but what about the angle? as always, any help would be awesome.

2. Jan 8, 2005

### vincentchan

Hints:
$$\tan{angle} = y(t)/100$$
And you wanna know d(angle)/dt at t=2

Last edited: Jan 8, 2005
3. Jan 8, 2005

### ziddy83

ok, correct me if im wrong (which i think i am)...so to relate the two, the height and the angle, i can use the following function, $$arctan (\frac {60t-5t^2} {100})$$
and then differentiate that to find the rate of change of the angle, right? :uhh:

4. Jan 8, 2005

### vincentchan

yeap, you got it, but remember the angle is in radian instead of degree

5. Jan 8, 2005

### ziddy83

great...thanks alot you guys