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An injured monkey sits perched on a tree branch 9 m above the ground, while a wildlife veterinarian is kneeling down in the bushes 90.0 m away attempting to subdue the monkey with a tranquilizer gun. The vet knows that the moment the gun fires, the monkey will be frightened and fall down from the branch. At what angle up from the ground must the veterinarian aim the gun so that the tranquilizer dart will hit the falling monkey? Given the angle, what is minimum speed at which the tranquilizer dart must leave the gun to still hit the monkey?

Respuesta :

Answer:

The hunter should aim directly at the perched monkey because the tranquilizer dart will fall away from the line sight at the same rate that the monkey falls from its perch.

Tan theta = 9 / 90 = .1      so theta = 5.71 deg

The time for the monkey to reach the ground is

t = (2 h / g)^1/2 = (18 / 9.8)^1/2 = 1.36 sec

So the horizontal speed of the dart must be at least

Vx = 90 m / 1.36 sec = 66.4 m/s

Vx = V cos theta

V = 66.4 m/s / cos 5.71 = 66.7 m/s

onyebuchinnaji

The angle of projection of the tranquilizer dart is 5.7⁰

The horizontal speed of the tranquilizer dart is 66.5 m/s

The given parameters;

  • vertical position of the monkey, y = 9 m
  • horizontal position of the veterinarian, x = 90

The angle of projection of the tranquilizer dart is calculated as;

[tex]tan(\theta) = \frac{y}{x} \\\\tan(\theta) = \frac{9}{90} \\\\tan(\theta) = 0.1\\\\\theta = tan^{-1}(0.1)\\\\\theta = 5.71 ^0[/tex]

The speed of the tranquilizer dart is calculated as;

X = v₀ₓt

where;

t is the time to reach maximum height

X is the horizontal displacement = 90 m

The time to reach maximum height is calculated as;

[tex]h = v_0_yt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 9}{9.8} } \\\\t = 1.36 \ s[/tex]

The horizontal speed of the tranquilizer dart is calculated as;

[tex]X = v_0cos(\theta) \times t\\\\90 = v_0\times cos(5.71) \times 1.36\\\\90 = 1.353 v_0\\\\v_0= \frac{90}{1.353} \\\\v_0 = 66.5 \ m/s[/tex]

Learn more here: https://brainly.com/question/20064545

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