Jason hits a volleyball with an initial velocity of 6 meters per second straight up. if the back starts at 2 meters above the floor how long will the ball be in the air

Answers

Answer 1

Answer: Vi = 6 m/s
d = -2 m
a = - 9.81 m/s^2
t = ?
Formula
d = vi*t + 1/2 a t^2
- 2 = 6m/s *t - .5 * 9.81 * t^2
Put the - 2 on the right.
4.905 t^2 + 6 m/s t +2 = 0
a = -4.905 m/s^2
b = 6 m/s
c = 2 m
t = [- b +/- square root(b^2 - 4ac)]/(2a)
t = [ -6 +/- square root (6^2 - 4*(-4.905)*2)] / (- 4.905*2)
t = [-6 +/- 8.674) / - 9.81
t = -14.674/-9.81
t = 1.69 seconds.

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Answers

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A train traveling at 82.8 kilometers per hour is blowing its horn as it approaches you. The horn has a frequency of 312 Hz. Assume the speed of the sound is 331 m/s. What is the observed frequency of the horn?335 Hz
312 Hz
292 Hz
412 Hz

Answers

The answer is 335 Hz.

To calculate this, we will use the following formula:
f₀ = f × v ÷ (v - v₁)
where:
f₀ - frequency heard
f - actual frequency
v - speed of sound
v₁ - speed of source

We know:
f = 312 Hz
v = 331 m/s
v₁ = 82.8 km/h = 23 m/s

We need:
f₀ = ?

f₀ = f × v ÷ (v - v₁)
⇒ f₀ = 312 Hz x 331 m/s ÷ (331 m/s - 23 m/s)
⇒ f₀ = 312 Hz × 331 m/s ÷ 308 m/s
⇒ f₀ = 335.3 Hz ≈ 335 Hz

Thus, the frequency of the horn is 335 Hz.

Answer:

335 Hz

Explanation:

100%

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Answers

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Answers

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