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Ilya [14]
2 years ago
14

a plane flying at a certain altitude is observed from two points that are 2.2 meters apart. the angles of elevation made by the

two points are 55º and 72º. if the plane is to the east of both observation posts, the altitude of the plane is . nextreset
Mathematics
1 answer:
IceJOKER [234]2 years ago
6 0
When you make a drawing with the instructions, and call x the altitude of the plane and y the distance from the angle 72 ° to the point straight below the plane, you obtain these equations

tan(55°) = x / (2.2 + y)

tan (72°) = x / y

You solve y from the second equation

y = x / tan(72°)

Substitute y in the first equation

tan (55°) = x /(2.2 + x/tan(72°)

Solve for x:

x = tan(55)[2.2 + x/tan(72) ]

x = tan(55) (2.2) + x tan(55)/tan(72)

x - x tan (55)/tan(72) = 2.2 tan (55)

Now substitute the values of the tangents

tan (55) = 1.4281
tan(72) = 3.0777

x - 0.464x = 3.1419

x = 3.1419 / (1-0.464) =5.86 m

Ah ... it is a toy plane! I guess.
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Kesha threw her baton up in the air from the marching band platform during practice. The equation h(t) = −16t² + 54t + 40 gives
lapo4ka [179]

Answer:

a) 40 feet

b) 54 ft/min

c) 4 mins

Step-by-step explanation:

Solution:-

- Kesha models the height ( h ) of the baton from the ground level but thrown from a platform of height hi.

- The function h ( t ) is modeled to follow a quadratic - parabolic path mathematically expressed as:

                           h ( t ) = −16t² + 54t + 40

Which gives the height of the baton from ground at time t mins.

- The initial point is of the height of the platform which is at a height of ( hi ) from the ground level.

- So the initial condition is expressed by time = 0 mins, the height of the baton h ( t ) would be:

                         h ( 0 ) = hi = -16*(0)^2 + 54*0 + 40

                         h ( 0 ) = hi = 0 + 0 + 40 = 40 feet

Answer: The height of the platform hi is 40 feet.

- The speed ( v ) during the parabolic path of the baton also varies with time t.

- The function of speed ( v ) with respect to time ( t ) can be determined by taking the derivative of displacement of baton from ground with respect to time t mins.

                        v ( t ) = dh / dt

                        v ( t )= d ( −16t² + 54t + 40 ) / dt

                        v ( t )= -2*(16)*t + 54

                        v ( t )= -32t + 54

- The velocity with which Kesha threw the baton is represented by tim t = 0 mins.

Hence,

                        v ( 0 ) = vi = -32*( 0 ) + 54

                        v ( 0 ) = vi = 54 ft / min

Answer: Kesha threw te baton with an initial speed of vo = 54 ft/min

- The baton reaches is maximum height h_max and comes down when all the kinetic energy is converted to potential energy. The baton starts to come down and cross the platform height hi = 40 feet and hits the ground.

- The height of the ball at ground is zero. Hence,

                     h ( t ) = 0

                     0 = −16t² + 54t + 40

                     0 = -8t^2 + 27t + 20

- Use the quadratic formula to solve the quadratic equation:

                     

                    t = \frac{27+/-\sqrt{27^2 - 4*8*(-20)} }{2*8}\\\\t = \frac{27+/-\sqrt{1369} }{16}\\\\t = \frac{27+/-37 }{16}\\\\t =  \frac{27 + 37}{16} \\\\t = 4

Answer: The time taken for the baton to hit the ground is t = 4 mins

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2√3/√3 = 2

So the side opposite  of the 30° = 2

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Answer:

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