A boy throws an arrow at an original velocity of 2m / s, aiming to create an angle 0, referring to the balloon at a distance of 3m from the point of departure. Calculate the angle 0 and the height of the arrow.

Answers

Answer 1

Answer:

s=0.204m

Explanation:

Assuming that the arrow is thrown horizontally and there is no air resistance, we can use the following formula to calculate the launch angle 0:

tan(0) = opposite/adjacent = height/distance

where opposite is the height that the arrow reaches and adjacent is the distance to the balloon.

Rearranging the formula, we get:

0 = arctan(height/distance)

0 = arctan(height/3)

Taking the tangent of both sides, we get:

tan(0) = tan(arctan(height/3))

tan(0) = height/3

Now, we need to find the height of the arrow. Using the kinematic equation:

v^2 = u^2 + 2as

where v is the final velocity (0m/s, at maximum height), u is the initial velocity (2m/s), a is acceleration (-9.8m/s^2, due to gravity) and s is the distance travelled vertically until the arrow reaches maximum height.

At maximum height, the final velocity is 0m/s. Therefore, we have:

0 = (2m/s)^2 + 2(-9.8m/s^2)s

Solving for s, we get:

s = 0.204m

Therefore, the height of the arrow is approximately 0.204m.

Related Questions

Answer the following Critical Thinking Questions. Each answer is worth 5 points, for a total of 25 points.

A particular star is 20 pc away from the Earth, and its luminosity is 160 times the luminosity of the Sun and has a surface temperature of 4000 K. Its absolute magnitude is -0.66. The temperature of the Sun is 5800 K. Explain/show your work.

What is this star's parallax?
What is this star's spectral class?
What is the wavelength at which this star radiates the most energy?
What is this star's apparent magnitude?
What is this star's radius, in solar radii?

Answers

The distance to the star in parsecs is given as 20 pc.

Using the absolute magnitude (M) and apparent magnitude (m) relation, we can find the star's apparent magnitude:

m - M = -5 + 5 log(d)

where d is the distance to the star in parsecs.

Plugging in the values we have, we get:

m - (-0.66) = -5 + 5 log(20)

m = 3.34

Therefore, this star's apparent magnitude is 3.34.

The star's luminosity is 160 times that of the Sun.

Using the Stefan-Boltzmann law, we can find the star's radius:

L = 4πR²σT⁴

where L is the luminosity, R is the radius, σ is the Stefan-Boltzmann constant, and T is the surface temperature.

We can write the ratio of the star's luminosity to that of the Sun as:

L/Lsun = (R/Rsun)²(T/Tsun)⁴

Plugging in the values we have, we get:

160 = (R/Rsun)²(4000/5800)⁴

Solving for R, we get:

R = 10.7 R⊙

Therefore, this star's radius is 10.7 times that of the Sun.

Using Wien's law, we can find the wavelength at which the star radiates the most energy:

λmax = 2.898 × 10⁶ / T

Plugging in the values we have, we get:

λmax = 724.5 nm

Therefore, this star radiates most of its energy at a wavelength of 724.5 nm.

The star's surface temperature is 4000 K.

Using the Harvard spectral classification system, we can find the star's spectral class based on its surface temperature:

O B A F G K M
50,000 10,000 7500 6000 5200 3700 2400

The star's surface temperature falls in the range of a K-type star.

Therefore, this star's spectral class is K.

Finally, we can use the definition of parallax to find the star's parallax:

p = 1/d

where p is the parallax in arcseconds and d is the distance to the star in parsecs

A student plucks a guitar string and the vibrations produce a sound wave with a frequency of 650 hertz. The sound wave produced can best be described as a

Answers

Answer:

I think its could be C

Explanation:

I think c because it makes the most sense and seems logical

Which of the following are examples of light behaving like an electromagnetic wave? Choose all that apply.

Group of answer choices

Compton scattering

interference through two slits

diffraction through one slit

photoelectric effect

refraction

Answers

The examples of light behaving like an electromagnetic wave are Compton scattering, interference through two slits, diffraction through one slit, and refraction.

Light is a type of electromagnetic radiation that is composed of electromagnetic waves. Light behaves like an electromagnetic wave in several ways. Electromagnetic waves are transverse waves that travel through a vacuum. They don't need a medium to propagate. Light can behave like an electromagnetic wave in several ways. Let's discuss each option in the question.
Compton scattering: Compton scattering is a phenomenon in which an incident X-ray or gamma-ray photon collides with an electron resulting in a scattered photon and a recoiling electron. It can only be explained by assuming that light behaves as both waves and particles. Therefore, Compton scattering is an example of light behaving like an electromagnetic wave.
Interference through two slits: When light passes through two narrow slits separated by a distance that is small compared to the wavelength of the light, it will diffract and interfere. The interference pattern will be characterized by bright and dark fringes. This phenomenon is an example of light behaving like an electromagnetic wave.
Diffraction through one slit: When light passes through a single narrow slit, it diffracts and creates an interference pattern similar to that produced by two slits. This phenomenon is an example of light behaving like an electromagnetic wave.
Photoelectric effect: The photoelectric effect is a phenomenon in which electrons are ejected from a metal surface when light shines on it. The photoelectric effect can be explained by assuming that light behaves as a stream of particles or photons. Therefore, the photoelectric effect is not an example of light behaving like an electromagnetic wave.
Refraction: Refraction is the bending of light as it passes from one medium to another, such as from air to water. It can be explained by assuming that light behaves like an electromagnetic wave. Therefore, refraction is an example of light behaving like an electromagnetic wave.

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Using a suitable diagram illustrate the following as applicable to the wave . 1. crest 2. Trough 3. Amplitude 4.Wavelength 5.Period

Answers

The images that shows the terms have been attached to this answer.

What does the terms mean?

A crest refers to the highest point or peak of a wave. It represents the maximum positive displacement or upward excursion of the wave from its rest position.

A trough is the maximum negative displacement or downward excursion of the wave from its rest position.

The amplitude of a wave refers to the maximum displacement or distance from the rest position to either the crest or the trough.

The wavelength is the distance between two adjacent crests or two adjacent troughs. In other words, it is the length of one complete wave cycle.

Period is the the duration between two corresponding points in a wave, such as two adjacent crests or troughs passing a fixed point.

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