PLEASE HELP MEEEE THIS IS DUE IN TEN MINUTES ALSO HOW THE HECK DO YOU CHANGE THE THING THAT SAYS COLLEGE BY UR NAME? IM IN FLIPP
ING EIGTH GRADE NOT COLLEGE!!!
2 answers:
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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°
Lets, assume the height(AB) = h and base(BC)= x
If the angle of elevation, ∠ACB = α, then
tan(α) =
Taking inverse trigonometric function, α = tan⁻¹ () .............(1)
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) :
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 = 10 feet/second.
C is the observer's position which is 50 feet away from the point B, so x = 50 feet.
= 0.12 (Rounding up to two decimal places)
So, the rate of change of the angle of elevation is 0.12 radians/second.
