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3 years ago

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She sights two sailboats going due east from the tower. The angles of depression to the two boats are 42o and 29o. If the observ

ation deck is 1,353 feet high, how far apart are the boats?

1 answer:

Reika [66]3 years ago

3 0

Answer:

The boats are 934.65 feet apart

Explanation:

Given:

The angles of depression to the two boats are 42 degrees and 29 degrees

Height of the observation deck i = 1,353 feet

To Find:

How far apart are the boats (y )= ?

Solution:

<em><u>Step 1 : Finding the value of x(Refer the figure attached)</u></em>

We can use the tangent ratio to find the x value

$tan(42^{\circ}) = \frac{1353}{x}$

$x = \frac{1353}{tan(42^{\circ}) }$

x = 590.47 feet

<em><u>Step 2 : Finding the value of z (Refer the figure attached)</u></em>

$tan(29^{\circ}) = \frac{1353}{z }$

$z = \frac{1353}{tan(29^{\circ})}$

z = 1525.12 feet

<em><u>Step 3 : Finding the value of y (Refer the figure attached</u></em>)

y = z -x

y = 1525.12 - 590.47

y = 934.65 feet

Thus the two boats are 934.65 feet apart

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2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

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$s$ - Final height, measured in meters.

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The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

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$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

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