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Alekssandra [29.7K]
3 years ago
6

Need help thanks ---

Mathematics
1 answer:
Molodets [167]3 years ago
6 0

the answer is A

these are filler words so I can submit it haha

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The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

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If the angle of elevation, ∠ACB = α, then

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As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and \frac{dh}{dt} = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\   \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512

= 0.12 (Rounding up to two decimal places)

So, the rate of change of the angle of elevation is 0.12 radians/second.

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