Angle of line of sight for a vertically lauched rocket

AI Thread Summary
The discussion focuses on calculating the rate at which an observer must change the angle of elevation to keep a vertically launched rocket in view. The rocket's height is modeled by the equation y(t) = 60t - 5t^2, and the observer is positioned 100 meters away horizontally. After two seconds, the derivative of the height, y', is used to determine the rate of change of height. The relationship between height and angle of elevation is established using the arctan function, leading to the differentiation needed to find the rate of change of the angle. The importance of using radians for angle measurement is also emphasized.
ziddy83
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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 which the height is changing, but what about the angle? as always, any help would be awesome.
 
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Hints:
\tan{angle} = y(t)/100
And you want to know d(angle)/dt at t=2
 
Last edited:
ok, correct me if I am 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? :rolleyes:
 
yeap, you got it, but remember the angle is in radian instead of degree
 
great...thanks a lot you guys
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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