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erastovalidia [21]
3 years ago
14

A hot air balloon rising vertically is tracked by an observer located 3 miles from the lift-off point. At a certain moment, the

angle between the observer's line-of-sight and the horizontal is π3π3 , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment?
Physics
1 answer:
yuradex [85]3 years ago
5 0

Answer:

\frac{dy}{dt}=1.2\frac{mi}{min}

Explanation:

We know that the tangent function relates the angle of the right triangle that forms the hot air balloon rising:

tan\theta=\frac{y}{x}\\y=xtan\theta(1)

Differentiating (1) with respect to time, we get:

\frac{dy}{dt}=tan\theta\frac{dx}{dt}+xsec^{2}\theta\frac{d\theta}{dt}\\

\frac{dx}{dt}=0 since x is a constant value. Replacing:

\frac{dy}{dt}=3mi(sec^{2}\frac{\pi}{3})0.1\frac{rad}{min}\\\frac{dy}{dt}=1.2\frac{mi}{min}

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Answer:

c. 4.582\times10^{21} kg

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r_{i} = Initial distance between asteroid and rock = 7514 km = 7514000 m

r_{f} = Final distance between asteroid and rock = 2823 km = 2823000 m

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Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg

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