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Assoli18 [71]
2 years ago
9

A stranded soldier shoots a signal flare into the air to attract the attention of a nearby plane. The flare has an initial verti

cal velocity of 1500 feet per second. Its height is defined by the quadratic function below. Assume that the flare is fired from ground level. h=v1t-16t^2
1. What is the maximum height that the flare reaches?
2. When will the flare reach that height?
3. At what time does the flare hit the ground again?
4. If the plane is flying at a height of 30,000 feet, a speed of 880 feet per second and is 50,000 feet from the fare when it is fired, will the flare hit it? If so, tell when this will happen. If not, tell when the flare reaches the planes altitude.
Physics
1 answer:
Ipatiy [6.2K]2 years ago
8 0

Answer:

Explanation:

h = ut - 16 t² = ut - 1/2 x32 t² = ut - 1/2 g t² , g = acceleration  = - 32 ft / s²

1) v² = u² - 2 g h , v = 0 so

h = u² / 2g = 1500² / 2 x 32 = 35156.25 ft

2) v = u - gt

t = u / g = 1500 / 32 = 46.875 s

3) It will hit the ground after 2 x 46.875 = 93.75 s

4 ) time to reach 30000 ft height  t is given by

h = ut - 16 t²

30000 = 1500t - 16t²

16t²-1500t + 30000 = 0

t = 28.92 s  and 64.82 s

Time required to travel 50000 by plane

= 50000/880 = 56.82 . There is no match of timing so plane will not hit it.

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A 70 kg student jumps down to form a 1 m high platform. She forgets to bend her knees and her downward motion stops in 0.02 seco
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Answer:

15,505 N

Explanation:

Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student

-ΔU = ΔK

-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²

-(0 - mgh) = 1/2mv² - 0

mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform  = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.

mgh = 1/2mv²

gh = 1/2v²

v² = 2gh

v = √(2gh)

v = √(2 × 9.8 m/s² × 1 m)

v = √(19.6 m²/s²)

v = 4.43 m/s

Upon impact on the ground and stopping, impulse I = Ft = m(v' - v) where F = force, t = time = 0.02 s, m =mass of student = 70 kg, v = initial velocity on impact = 4.43 m/s and v'= final velocity at stopping = 0 m/s

So Ft = m(v' - v)

F = m(v' - v)/t

substituting the values of the variables, we have

F = 70 kg(0 m/s - 4.43 m/s)/0.02 s

= 70 kg(- 4.43 m/s)/0.02 s

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2 years ago
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Betelgeuse is the bright red star representing the left shoulder of the constellation Orion. All the following statements about
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Answer:

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2 years ago
Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another
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Answer:

6.1875\times 10^{-8}

Explanation:

Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies I=\frac {k}{r^{2}} where r is the distance

Since intensity is given then when at 3 m

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