Chapter 4: Q41P (page 86)

Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water (Figure). Although the fish sees the insect along a straight-line path at angle $ϕ$ and distance d , a drop must be launched at a different angle $θ_{0}$ if its parabolic path is to intersect the insect. If $ϕ = 36 °$ and $d = 0.900 m$ , what launch angle $θ_{0}$ is required for the drop to be at the top of the parabolic path when it reaches the insect?

Short Answer

Expert verified

The launch angle $θ_{0} = 55 . 5^{o}$

Step by step solution

Given information

It is given that,

$ϕ = 36 . 0^{o}$

Determining the concept

The problem deals with the projectile motion of an object. The projectile motion of an object is the motion of an object thrown into the air, subject to the acceleration due to gravity Though the drop is launched at a different angle, its launching point is the same as that of the archerfish. So, they have to travel the same distance. Using this, the angle can be found.

Formulae:

The range of the projectile is given by,

$R = \frac{(v_{0}^{2} \sin n 2 θ_{0})}{g}$

Maximum height is given by,

$H_{m a x} = \frac{(v_{0}^{2} s i n^{2} θ_{0})}{2 g}$

Determining the launch angle θ0

From the diagram,

$Y = H = \frac{(v_{0}^{2} \sin^{2} θ_{0})}{2 g} x = \frac{R}{2} = \frac{(v_{0}^{2} \sin 2 θ_{0})}{2 \times g} \tan ϕ = \frac{Y}{H} \tan ϕ = \frac{\frac{(v_{0}^{2} \sin^{2} θ_{0})}{2 g}}{\frac{(v_{0}^{2} \sin 2 θ_{0})}{2 \times g}} \tan ϕ = \frac{1}{2} \times \frac{\sin θ_{0}}{\cos θ_{0}} \tan ϕ = \frac{1}{2} \times \tan θ_{0} \tan ϕ = 2 \times \tan ϕ θ_{0} = \tan^{- 1} (2 \times \tan ϕ) θ_{0} = \tan^{- 1} (2 \times \tan 36.0) θ_{0} = \tan^{- 1} (1.453)$

Thus,

$θ_{0} = 55.46 \approx 55.5 °$

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Short Answer

Step by step solution

Given information

Determining the concept

Determining the launch angle θ0

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