1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Luda [366]
3 years ago
15

Can someone please help a struggling physics student?

Physics
1 answer:
zhenek [66]3 years ago
5 0
<h3><u>Part A:</u></h3>

<u><em>What is the maximum height the ball will reach in the air?</em></u>

Kinematics equation used:

  • v_f^2=v_i^2+2ad, where v_f is final velocity, v_i is initial velocity, a is acceleration, and d is distance travelled. From SI units, velocity should be in m/s, acceleration should be in m/s^2, and distance should be in m

We're given that the initial velocity is 12.0 m/s in the y-direction. At the maximum height, the vertical velocity of the ball will be 0 m/s, otherwise it would not be at maximum height. This is our final velocity.

The only acceleration in the system is acceleration due to gravity, which is approximately 9.8\:\mathrm{ m/s^2}. However, the acceleration is acting down, whereas the ball is moving up. To express its direction, acceleration should be plugged in as -9.8\:\mathrm{m/s^2}. We have three variables, and we are solving for the fourth, which is distance travelled. This will be the maximum height of the ball.

Substitute v_i=12, v_f=0, a=-9.8 to solve for d:

0^2=12^2+2(-9.8)(d),\\0=144-19.6d,\\-19.6d=-144,\\d=\frac{-144}{-19.6}=7.34693877551\approx \boxed{7.35\text{ m}}

<u><em>What is the velocity of the ball when it hits the ground?</em></u>

This question tests a physics concept rather than a physics formula. The vertical velocity of the ball when it hits the ground is equal in magnitude but opposite in direction to the ball's initial vertical velocity. This is because the ball spends equal time travelling to its max height as it does travelling from max height to the ground (ball is accelerating from initial velocity to 0 and then from 0 to some velocity over the same distance and time). Since the ball has an initial vertical velocity of +12.0 m/s, its velocity when it hits the ground will be \boxed{-12.0\text{ m/s}}. (The negative sign represents the direction. Because velocity is a vector, it is required.)

<h3><u>Part B:</u></h3>

<u>**Since my initial answer exceeds the character limit, I've attached the first question to Part B as an image. Please refer to the attached image for the answer and explanation to the first question of Part B. Apologies for the inconvenience.**</u>

<u><em>What is the direction of the velocity of the ball when it hits the ground? Express your answer in terms of the angle (in degrees ) of the ball's velocity with respect to the horizontal direction (see figure).</em></u>

This question uses a similar concept as the second question of Part A. The vertical velocity of the ball at launch is equal in magnitude but opposite in direction to the ball's final velocity. The horizontal component is equal in both magnitude and direction throughout the entire launch, since there are no horizontal forces acting on the system. Therefore, the angle below the horizontal of the ball's velocity when it hits the ground is equal to the angle of the ball to the horizontal at launch.

To find this, we need to use basic trigonometry for a right triangle. In any right triangle, the tangent/tan of an angle is equal to its opposite side divided by its adjacent side.

Let the angle to the horizontal at launch be \theta. The angle's opposite side is represented by the vertical velocity at launch (12.0 m/s) and the angle's adjacent side is represented by the horizontal velocity at launch (2.3 m/s). Therefore, we have the following equation:

\tan \theta=\frac{12.0\text{m/s}}{2.3\text{ m/s}}

Take the inverse tangent of both sides:

\arctan (\tan \theta)=\arctan (\frac{12.0}{2.3})

Simplify using \arctan(\tan \theta)=\theta \text{ for }\theta \in (-90^{\circ}, 90^{\circ}):

\theta=\arctan(\frac{12.3}{2.3}),\\\theta =79.14989537\approx \boxed{79.15^{\circ}}

We can express our answer by saying that the direction of the velocity of the ball when it hits the ground is \boxed{\text{approximately }79.15^{\circ} \text{ below the horizontal}} or \boxed{\text{approximately }-79.15^{\circ} \text{ to the horizontal}}.

You might be interested in
Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in
postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is \rho = \frac{m}{V}, and the volume of a pyramid is (confusingly written on the proposal) V=\frac{1}{3} Ah, so we can write:

m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h

Where s is the side of the base, being s^2 the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that \frac{1m}{100cm}=1, and we can use this factor to convert anything written in cm to anything written in m. Example:

500cm=500cm\frac{1m}{100cm}=5m

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3

Where we have used the fact that 1^3=1 (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg

And now we will convert to short tons using two conversion factors at the same time:

m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0
2 years ago
You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g&#10;            y- 0 = 10.0²/2 9.8&#10;            y - 0 = 5.10 m&#10;            &#10;The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system&#10;             y₂ = 5.1 + 44&#10;             y₂ = 49.1 m&#10;Let's use the other equation to find the time&#10;              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

7 0
3 years ago
How much energy goes toward kinetic energy
kirill [66]

kinetic is moving

so kinetic energy is something that moves

8 0
3 years ago
Read 2 more answers
In an internal combustion engine, heat flow into a gas causes it to ......
ivann1987 [24]
<span>In an internal combustion engine, heat flow into a gas causes it to expand.
The application of direct force to specific parts of the engine will produce </span>expansion of the high-temperature<span> and high-</span>pressure<span> gases. Which will transform the chemical energy from the fuel (such as gasoline or oi) into mechanical energy.</span>
5 0
3 years ago
Two cars collide at an intersection. Car A , with a mass of 2000 kg, is going from west to east, while car B, of mass 1500 kg, i
devlian [24]

Answer:

The answer of the part (a) is v2 = 7.09 m/s

and the answer of the part (b) is vA1 = 5.25 m/s

Explanation:

Explanation of the both parts of answer  is in the following attachments

6 0
3 years ago
Other questions:
  • Select the correct answer.
    5·1 answer
  • How many centimeters are in 4 inches?
    10·1 answer
  • A mechanic uses a hydraulic car jack to lift the front end of a car to change the oil. The jack used exerts 8,915 N of force fro
    6·1 answer
  • Estimate the number of dollar bills (15.5 cm wide), placed end to end, that it would take to circle the Earth (radius = 6.40 × 1
    7·1 answer
  • Light waves from the sun can be converted to electricity through __________.
    7·1 answer
  • 2 (a) What is the distance from the Sun to Earth in terms of solar radii? Earth radii?
    12·1 answer
  • A 63.0-kg man is riding an escalator in a shopping mall. The escalator moves the man at a constant velocity from ground level to
    6·1 answer
  • You go to the playground and slide down the slide, a 3.6-m-long ramp at an angle of 40. ∘ with respect to horizontal. The pants
    10·1 answer
  • Write relationship between hertz and megahertz​
    13·1 answer
  • If you are given force and time, you can determine power if you can know
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!