Answer :
Answer:
v = 21.6 Km/min
Step-by-step explanation:
x = 6 Km
∅ = π/3 (variable)
d∅/dt = 0.9 rad/min
Let ∅ be the angle between the spy's line of sight and the ground and y is the distance between the rocket and the ground.
Then we have the equation
tan ∅ = y / x
⇒ (tan ∅)' = (y / x)' = (1/x)*y'
⇒ Sec²∅*(d∅/dt) = (1/x)*(dy/dt) = (1/x)*v
⇒ v = x*Sec²∅*(d∅/dt)
⇒ v = 6 Km*Sec²(π/3)*(0.9 rad/min)
⇒ v = 21.6 Km/min
Based on the calculations, the rocket’s velocity at this moment is equal to 21.6 km/min.
Given the following data:
- Vertical distance = 6 kilometers.
- Angle of inclination = [tex]\frac{\pi}{3}[/tex]
- Rate of change = 0.9 rad/min.
How to calculate rocket’s velocity.
First of all, we would assign a variable to the given parameters. So, let the angle formed between the ground and the spy's line of sight be [tex]\theta[/tex] while the vertical distance between the rocket and the ground would be y.
From Tan trigonometric function, we have:
[tex]Tan \theta =\frac{y}{x}[/tex]
[tex]Tan\theta' = (\frac{y}{x} )'= \frac{1}{x} \times y'[/tex]
Note: In a right-angled triangle, the secant (sec) of an angle is equal to the ratio of the length of the hypotenuse to the length of the adjacent side.
[tex]Sec^2\theta(\frac{d\theta }{dt} )=\frac{1}{x} (\frac{dy}{dt} )=\frac{1}{x}(v)=\frac{v}{x}[/tex]
Cross-multiplying, we have:
[tex]V= xSec^2\theta(\frac{d\theta }{dt} )\\\\V= 6 \times Sec^2(\frac{\pi }{3} )\times 0.9\\\\V= 6 \times Sec^2(\frac{180 }{3} )\times 0.9[/tex]
Velocity, V = 21.6 km/min.
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