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Ulleksa [173]
3 years ago
13

The light beacon of a lighthouse is situated a distance 2.00 km from a long, straight shoreline (shortest distance to the shorel

ine). The light rotates at a rate of 3.00 rev/minute. At a certain time, the illuminated spot on the shoreline is 4.00 km from the lighthouse. How fast is the illuminated spot moving at that point?

Physics
1 answer:
kolbaska11 [484]3 years ago
6 0

In triangle ABC

tanθ = BC/AB

tanθ = x/2

taking derivative both side relative to "t"

sec²θ (dθ/dt) = (0.5) (dx/dt)

sec²θ w = (0.5) v                                eq-1

when x = 4 ,

tanθ = 4/2

θ = tan⁻¹ (2) = 63.4 deg

using eq-1

sec²θ w = (0.5) v  

sec²63.4 (3) (2pi/60) = (0.5) v  

v = 3.13 m/s

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Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

Explanation:

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r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

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With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

y = y0 + v0y · t + 1/2 · g · t²

Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

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