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Elena-2011 [213]
3 years ago
6

A radio station tower was built in two sections. From a point 87 feet from the base of the tower, the angle of elevation of the

top of the first section is 25 degrees, and the angle of the top of the second section is 40 degrees. to the nearest foot, what is the height of the top section of the tower?
Physics
1 answer:
Fiesta28 [93]3 years ago
3 0

Answer:

tan theta = H / B        where H is height of triangle and B the length of base

H1 = 87 * tan 25 =  40.6 feet

H2 = 87 * tan 40 = 73 feet

H2 - H1 = 32.4 feet

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Two stunt drivers drive directly toward each other. At time t=0 the two cars are a distance D apart, car 1 is at rest, and car 2
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Answer: Hello there!

We know this:

The distance between the cars at t= 0 is D.

car 2 has an initial velocity of v0 and no acceleration.

car 1 has no initial velocity and a acceleration of ax that starts at  t = 0

then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.

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and in this way we could ignore constants of integration :D

for the position of each car we integrate again:  

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now we can solve it for t using the Bhaskara equation.

t = \frac{-v0 +\sqrt{v0^{2} + 4*(1/2)ax*D } }{2(1/2)ax} =\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}

that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).

Now we want to know the velocity of car 1 just before the impact, this can be calculated by valuating the time in the as the time that we just found in the velocity equation for the car 1, this is:

v(\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}) = ax*\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax} = {-v0 +\sqrt{v0^{2} + 2ax*D }

where again, you need to replace the values of v0, D and ax.

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3 years ago
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Answer:

The options are not shown, so let's derive the relationship.

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So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.

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