A rocket rises vertically from a point on the ground that is 100m from an observer at ground level. The observer notes that the angle of elevation is increasing at a rate of 12∘per second when the angle of elevation is 60∘. Find the speed of the rocket at that instant. Hint: remember to use radians.
make a diagram.
you should have a right-angled triangle, where the angle of elevation is Ø, the adjacent is 100 and the opposite is h, the height of the rocket.
tanØ = h/100
h = 100tanØ
dh/dt = 100sec^2 Ø (dØ/dt)
when Ø = 60° or π/3 rad
dØ/dt = 12°/sec or 12π/180 rad/sec = π/15 rad/sec
sub into the dh/dt equation and you are done
Well, this seems like a rocket science question! But don't worry, I'm here to lighten the mood. So, let's get started with the problem.
First, let's convert the given angle of 60 degrees to radians. To do that, we'll use the conversion factor of π/180 since there are π radians in 180 degrees. So, 60 degrees equals 60 * π/180 radians, which simplifies to π/3 radians.
Now, let's call the distance between the rocket and the observer "x" and the angle of elevation "θ." We need to find the speed of the rocket, which is the rate of change of x with respect to time.
Let's differentiate the equation x = 100/(tanθ) with respect to time, which will give us the rate of change of x:
dx/dt = (d(100/tanθ))/dt
Since θ is the angle of elevation, we can express tanθ as tan(θ) = x/100.
So, our equation becomes:
dx/dt = (d(100/(x/100)))/dt
Simplifying further:
dx/dt = (d(100*100/x))/dt
= (d(10000/x))/dt
= (-10000/x^2) * dx/dt
Now, we're given that dθ/dt (the rate of change of the angle of elevation) is 12 degrees per second. We'll convert this to radians by multiplying it by π/180, which gives us dθ/dt = 12π/180 radians per second.
Since x = 100/(tanθ), we can substitute dx/dt and dθ/dt into our equation:
-10000/x^2 * dx/dt = 12π/180
Substituting x = 100/(tanθ):
-10000/(100/(tanθ))^2 * dx/dt = 12π/180
Simplifying:
-10000/(1/(tanθ))^2 * dx/dt = 12π/180
-10000/(1/(tanπ/3))^2 * dx/dt = 12π/180
Now, let's substitute tanπ/3 with √3:
-10000/(1/√3)^2 * dx/dt = 12π/180
Simplifying further:
-10000/(1/(3))^2 * dx/dt = 12π/180
-10000/(1/9) * dx/dt = 12π/180
-10000 * 9 * dx/dt = 12π/180
Finally, let's solve for dx/dt (the rate of change of x):
dx/dt = (12π/180) / (-10000 * 9)
dx/dt = π/135000
So, the speed of the rocket at that instant is π/135000 meters per second.
And there you have it! The speed of the rocket is undoubtedly out of this world. Keep reaching for the stars, my friend!
To solve this problem, we need to use trigonometry and the relationship between the angle of elevation and the distance between the observer and the rocket.
Let's denote the angle of elevation as θ and the distance between the observer and the rocket as x. The given information tells us that dx/dt = 12°/s and θ = 60°.
We can draw a right triangle to represent the situation. The vertical leg of the triangle represents the height of the rocket (h), and the horizontal leg represents the distance between the observer and the rocket (x), which is constant at 100m.
Now, let's use the trigonometric function tangent to relate the angle of elevation and the distance:
tan(θ) = h / x
Differentiating both sides with respect to time (t), we get:
sec^2(θ) * dθ/dt = dh/dt / x
Since θ = 60° and dx/dt = 12°/s, we can substitute these values into the equation:
sec^2(60°) * (12°/s) = dh/dt / 100m
To get the value of sec^2(60°), we need to convert the angle to radians:
60° * (π/180) = π/3 radians
Now, let's calculate the sec^2(60°):
sec(π/3) = 1 / cos(π/3) = 1 / 0.5 = 2
sec^2(60°) = (2)^2 = 4
Substituting this value back into the equation:
4 * (12°/s) = dh/dt / 100m
48°/s = dh/dt / 100m
Now, let's rearrange the equation to solve for dh/dt:
dh/dt = 48°/s * 100m
dh/dt = 4800 m/degree
Since we're given the speed in radians, we need to convert it:
dh/dt = 4800 m/degree * (π/180) rad/degree
dh/dt = 80π m/s
Therefore, the speed of the rocket at that instant is 80π m/s.
Note: π is approximately equal to 3.14159, so the final answer can be approximated to 251.33 m/s.
To find the speed of the rocket at that instant, we can make use of trigonometry and calculus.
Let's assume that at time t, the height of the rocket is h(t) meters, and the angle of elevation from the observer to the rocket is θ(t) degrees. We are given that the observer notes that the angle of elevation is increasing at a rate of 12 degrees per second when θ(t) is equal to 60 degrees.
To start, we need to convert the angle from degrees to radians. Since 1 radian is equal to 180/π degrees, we can convert the rate of change of the angle to radians per second by multiplying it by π/180:
dθ/dt = 12 * π/180 radians per second.
Next, let's consider the right triangle formed by the observer, the point on the ground, and the rocket. The height of the rocket, h(t), is the opposite side of the triangle, and the distance between the observer and the rocket, which is 100 meters, is the adjacent side.
We can use the tangent function to relate the angle θ(t) with the height of the rocket. The tangent of an angle is equal to the opposite side divided by the adjacent side:
tan(θ(t)) = h(t)/100.
To relate the rate of change of the angle with the rate of change of the height, we need to take the derivative of this equation with respect to time t:
sec^2(θ(t)) * dθ/dt = dh/dt / 100.
Since sec^2(θ(t)) is equal to 1 + tan^2(θ(t)), we can rewrite the equation as:
(1 + tan^2(θ(t))) * dθ/dt = dh/dt / 100.
At the instant where θ(t) is equal to 60 degrees, we have:
(1 + tan^2(60°)) * (12 * π/180) = dh/dt / 100.
Simplifying this equation, we can find dh/dt, the rate of change of the height:
dh/dt = (1 + (√3)^2) * (12 * π/180) * 100.
Finally, the speed of the rocket is equal to the rate of change of the height:
Speed = |dh/dt|.
Substituting the value for dh/dt obtained above, we can calculate the speed of the rocket.