The horizontal component of a projectiles velocity is independent of

1. The horizontal component of a projectiles velocity is independent of

Answer:

In the absence of air resistance, the horizontal component of a projectile's velocity does not change as the projectile moves. 7. At the instant a ball is thrown horizontally with a large force, an identical ball is dropped from the same height.

2. write true if the statement is correct and false if it is incorrect.1.a projectile is free-falling object2. a projectile is experience negligible or no air resistance3.the horizontal velocity of projectile changes by 9.8 m/s second4.the vertical component of the velocity of the projectile is constant5.the horizontal velocity of projectile is 0 m/s at the peak of its trajectory6.the final horizontal velocity of projectile is always equal to initial horizontal velocity7.a projectile with a horizontal component of motion will have a constant horizontal velocity8.the time that projectile is in the air dependent upon the horizontal component of the initial velocity9.the minimum speed of projectile is equal to horizontal component of the velocity of the projectile 10.a projectile rises towards the peak of its trajectory, the horizontal velocity will decrease; as it falls from the peak of its trajectory, its horizontal velocity will decrease​

Answer:

1.  true

2. true

3. true

4. false

5. true

6. false

7. true

8. false

9. true

10. true

Explanation:

hope it helps

3. why does the horizontal component of the projectile's velocity independent of time?NEEDED ASAP

Answer:

The horizontal motion of a projectile is independent of its vertical motion.

explanation: Because The horizontal velocity of a projectile is constant (a never changing in value), There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down, The vertical velocity of a projectile changes by 9.8 m/s each seco

4. A projectile is launched from and returns to ground level. Air resistance is absent. The horizontal range of the projectile is measured to be R = 175 m, and the horizontal component of the launch velocity is v0x = 25 m/s. Discuss briefly each answer on a. What is the final value of the horizontal component vx of the projectile’s velocity? b. Can the time be determined for the horizontal part of the motion? c. For the vertical part of the motion, what is the displacement of the projectile?​

Answer:

A projectile is launched from and returns to ground level, as Figure 3.22 shows. Air resistance is absent. The horizontal range of the projectile is R=175 m, and the horizontal component of the launch velocity is v0x=25 m/s. Find the vertical component v0y of the launch velocity.

A projectile, launched with a velocity whose horizontal component is

, has a range of

. From these data the vertical component v

0y

of the initial velocity can be determined.

Figure 3.22 A projectile, launched with a velocity whose horizontal component is , has a range of . From these data the vertical component v 0y of the initial velocity can be determined.

What is the final value of the horizontal component vx of the projectile’s velocity?

The final value vx of the horizontal component of the projectile’s velocity is the same as the initial value in the absence of air resistance. In other words, the horizontal motion occurs at a constant velocity of 25 m/s.

Can the time be determined for the horizontal part of the motion?

Yes. In constant-velocity motion, the time is just the horizontal distance (the range) divided by the magnitude of the horizontal component of the projectile’s velocity.

Is the time for the horizontal part of the motion the same as the time for the vertical part of the motion?

Yes. The value for the time calculated for the horizontal part of the motion can be used to analyze the vertical part of the motion.

For the vertical part of the motion, what is the displacement of the projectile?

Since the projectile is launched from and returns to ground level, the vertical displacement is zero.

Solution From the constant-velocity horizontal motion, we find that the time is

For the vertical part of the motion, we know that the displacement is zero and that the acceleration due to gravity is –9.80 m/s2, assuming that upward is the positive direction. Therefore, we can use Equation 3.5b to find the initial y component of the velocity:

Step-by-step explanation:

pa brainliest po ty

5. 1. Why does the vertical component of the projectile's velocity vary with time? 2. Why does the horizontal component of the projectile's velocity independent of time? 3. In the physical world, why do you think vertical and horizontal components of projectiles occur? Give examples 4. A baseball is thrown horizontally with an initial velocity. Describe its path and overall motion if there is no gravity 5. A baseball is thrown horizontally with an initial velocity. Describe its path and overall motion if there is gravity​

+++++++++++++++++++++

ANSWER;

1. Why does the vertical component of the projectile's velocity vary with time?

- Because the force of gravity on a projectile acts only vertically, hence only the vertical component of a projectile changes with time.

2. Why does the horizontal component of the projectile's velocity independent of time?

- Because The horizontal motion of a projectile is independent of its vertical motion.

3. In the physical world, why do you think vertical and horizontal components of projectiles occur? Give examples

- Because the horizontal and vertical motions of a projectile are independent, meaning they do not affect each other.

4. A baseball is thrown horizontally with an initial velocity. Describe its path and overall motion if there is no gravity

- Without gravity, the ball continues in a straight line at its initial velocity forever, or until it happens to collide with some other object in its path.

5. A baseball is thrown horizontally with an initial velocity. Describe its path and overall motion if there is gravity

- With gravity, the ball will drop vertically below its otherwise straight-line, inertial path.

+++++++++++++++++++++

6. The horizontal component of a projectile is independent of

Answer:

The vertical velocity of a projectile changes by 9.8 m/s each second, The horizontal motion of a projectile is independent of its

vertical motion

Answer:

Thd vertical velocity of a projectile changes by 9.8 m/s each second, The horizontal motion of a projectile is independent of its vertical motion

Explanation:

hope it helps

7. A projectile thrown upward at an angle theta with the horizontal covers a range of 12 cm in a span of 5.0 s. Find the (a) horizontal and vertical components of the initial velocity, (b) angle of projection theta, (c) initial velocity, and (d) maximum height attained by the projectile.​

(a) horizontal and vertical components of the initial velocity : 24.5 and 2.4 m/s

(b) angle of projection theta : 1.473°

(c) initial velocity : 24.618 m/s

(d) maximum height attained by the projectile : 30.625 m

Further explanation  

Projectile motion is a motion by particles that move along a curved path and are affected by gravity  

There are two motions in this projectile, horizontal motion and vertical motion  

Horizontal motion is called constant velocity motion with constant velocity and vertical motion is called uniformly accelerated motion with constant acceleration (a = g)  

a. horizontal and vertical components of the initial velocity,

total time :

[tex]\tt t=\dfrac{2.u.sin`\theta}{g}\\\\2.u.sin\theta=g.t\\\\u.sin\theta=\dfrac{9.8\times 5}{2}=24.5\rightarrow vertical~of~initial~velocity[/tex]

range :

[tex]\tt R=\dfrac{2u^2sin\theta~cos\theta}{g}\\\\R=\dfrac{2.u.sin\theta}{g}\times u.cos\theta\\\\12=\dfrac{2\times 24.5}{9.8}\times u.cos\theta\\\\u.cos\theta=\dfrac{12}{5}=2.4\rightarrow horizontal~of~initial~velocity[/tex]

b. angle of projection theta(θ)

[tex]\tt \dfrac{u.sin\theta}{u.cos\theta}=\dfrac{24.5}{2.4}\\\\tan\theta=10.2\\\\\theta=1.473^o[/tex]

c. initial velocity

[tex]\tt u.sin~1.473=24.5\\\\u=\dfrac{24.5}{sin~1.473}=24.618~m/s[/tex]

d. maximum height attained by the projectile.

[tex]\tt h_{max}=\dfrac{u^2sin^2\theta}{2g}\\\\h_{max}=\dfrac{(u.sin\theta)^2}{2.g}\\\\h_{max}=\dfrac{24.5^2}{2\times 9.8}\\\\h_{max}=30.625~m[/tex]

Learn more  

vertical motion, horizontal motion and projectile motion  

https://brainly.ph/question/1332444  

the horizontal and vertical motions of a projectile  

https://brainly.ph/question/2175047  

knowledge of projectile motion  

https://brainly.ph/question/1424636  

#LetsStudy  

8. All projectiles have uniform velocity along the horizontal component. True or False?

Answer:

False

Explanation:

A projectile has a constant horizontal velocity. The vertical velocity will change by 9.8 m/s each second..

9. Ill. Direction. Write TRUE if the statement is TRUE and FALSE if it is incorrect. 11 1. The horizontal motion of a projectile is independent of its vertical motion. 2. The horizontal velocity of a projectile is constant (a never changing in value). 3. There is a vertical acceleration caused by gravity; its value is 9.81 m/s/s, down. 4. The vertical velocity of a projectile changes by 9.81 m/s each second. 5. There is a horizontal forces acting upon projectiles and thus horizontal acceleration happens.​

Answer:

1)true2)true3)true4)true5)false

Explanation:

tama yan ace na ako

Answer:

1)True

2)True

3)True

4)True

5)True

Explanation:

I hope it helps you peoples...if its correct rate please and Like

10. Which of the following is independent of the horizontal component of a projectile velocity?a. its rangeb. the vertical componentc. timed. none of these​

Answer:

Which of the following is independent of the horizontal component of a

projectile velocity?

Answer:  >> D. None of These

Explanation: None of the above of the following is independent of the horizontal component of a  projectile velocity and that they are we don't  have influence on one another.

==========================================

#CarryOnLearning

The horizontal velocity of a projectile is constant (a never changing in value), There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down, The vertical velocity of a projectile changes by 9.8 m/s each second, The horizontal motion of a projectile is independent of its vertical motion.

11. A projectile is thrown with an initial velocity which has a horizontal component of 4 m/s. What will be its horizontal speed after 3 s?

Answer:

Horizontal distance traveled can be expressed as x = Vx * t , where t is the time. Vertical distance from the ground is described by the formula y = h + Vy * t – g * t² / 2 , where g is the gravity acceleration.

btw goodevening

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12. 1.A projectile is thrown with an initial velocity that has horizontal component of 4 m/s. What will be it's horizontal components of 4 m/s. What will be it's horizontal speed after 3 s?2. What is the value of the vertical speed at the highest point of the projectile's trajectory?pa help Naman Po guyss.. Yung maayos Po sana na sagot ​

Answer:

I assume that the body is a horizontally launched projectile coming from a higher elevation.

The horizontal velocity does not change with time but it's vertical velocity does. It accelerate at 9.8 m/s^2.

After 2 seconds, it's vertical velocity is already 19.6 m/s while it's horizontal velocity is still 4 m/s.

Solving for it's velocity using the Pythagorean theorem,

V^2 = Vx^2 + Vy^2

V^2 = 4^2 + 19.6^2

V^2 = 16 + 384.16

V^2 = 400.16

V = 20.00 m/s

It's final velocity after 2 seconds is 20.00 m/s.

Explanation:

13. The horizontal component of projectile has_____ velocitya.increasingb.decreasing c.constantd.all of the above ​

Answer:

D. ALL OF THE ABOVE

Explanation:

SANA MAKATULONG;)

GOOD NIGHT;)

14. write T on the blank if the statement is true and f it false.____1.Horizontal forces are not required to keep a projectile moving horizontally.____2.Gravity acts to influence the vertical motion of the projectile, this custing a vertical acceleration.____3.If the horizontal velocity of a projectile is constant, there is no horizontally acceleration.____4.The horizontal velocity of a projectile can be changed.____5.The horizontal displacement of a projectile is dependent upon the horizontal component of the initial velocity.​

Answer:

1.t

2.t

3.f

4.f

5.t

Explanation:

hope it help sa inyo

15. Given the angle and initial velocity, calculate the horizontal and vertical components of projectile​

Explanation:

blurred po sorry di koponmasagutan to

16. 1. A projectile is thrown with an initial velocity that has horizontal component of 4 m/s. What will be its horizontal components of 4 m/s. What will be its horizontal speed after 3 s? 2. What is the value of the vertical speed at the highest point of the projectile's trajectory?​

Answer:

1. 4 m/s

2. the value of the vertical speed at the highest point of the projectile's trajectory is the lowest speed at the maximum height reached.

Explanation:

#carryonlearning

17. A projectile is thrown with an initial velocity which has a horizontal component of 4m/s. What will be it's horizontal speed after 3s?

Answer:

I assume that the body is a horizontally launched projectile coming from a higher elevation.

The horizontal velocity does not change with time but it's vertical velocity does. It accelerate at 9.8 m/s^2.

After 2 seconds, it's vertical velocity is already 19.6 m/s while it's horizontal velocity is still 4 m/s.

Solving for it's velocity using the Pythagorean theorem,

V^2 = Vx^2 + Vy^2

V^2 = 4^2 + 19.6^2

V^2 = 16 + 384.16

V^2 = 400.16

V = 20.00 m/s

It's final velocity after 2 seconds is 20.00 m/s.

18. ACTIVITY Direction: Given the angle and initial velocity, calculate the horizontal and vertical components of projectile. Initial Angle (0) Velocity Horizontal Component Vertical Component 15° 10 m/s 9.66 m/s 2.59 m/s 30° 10 m/s 45° 10 m/s 60° 10 m/s 75° 10 m/s​

Answer:

Explanation:

The formulas we are going to use to compute for the horizontal and vertical components of initial velocity are

vix = vi cos θ

and

viy = vi sin θ

where

vix = horizontal component of velocity

viy = vertical component of velocity

vi = initial velocity

θ = angle of release (from the horizontal)

-----------------------------------------------------------

Solution: (30° and 10 m/s)

For the horizontal component

vix = vi cos θ

vix = (10 m/s)(cos 30°)

vix = (10 m/s)(0.866)

vix = 8.66 m/s

For the vertical component

viy = vi sin θ

viy = (10 m/s)(sin 30°)

viy = (10 m/s)(0.5)

viy = 5 m/s

-----------------------------------------------------------

Solution: (45° and 10 m/s)

For the horizontal component

vix = vi cos θ

vix = (10 m/s)(cos 45°)

vix = (10 m/s)(0.707)

vix = 7.07 m/s

For the vertical component

viy = vi sin θ

viy = (10 m/s)(sin 45°)

viy = (10 m/s)(0.707)

viy = 7.07 m/s

-----------------------------------------------------------

Solution: (60° and 10 m/s)

For the horizontal component

vix = vi cos θ

vix = (10 m/s)(cos 60°)

vix = (10 m/s)(0.5)

vix = 5 m/s

For the vertical component

viy = vi sin θ

viy = (10 m/s)(sin 60°)

viy = (10 m/s)(0.866)

viy = 8.66 m/s

-----------------------------------------------------------

Solution: (75° and 10 m/s)

For the horizontal component

vix = vi cos θ

vix = (10 m/s)(cos 75°)

vix = (10 m/s)(0.259)

vix = 2.59 m/s

For the vertical component

viy = vi sin θ

viy = (10 m/s)(sin 75°)

viy = (10 m/s)(0.966)

viy = 9.66 m/s

19. a projectile fired from the level ground has an initial vertical velocity component of 38 m/s and a horizontal velocity component 59 m/s.a. per second what is the total time of flight?b. the horizontal range of the projectile is closest to what meters?​

Answer:

Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s.

20. A projectile is fired with a horizontal velocity of 330 m/s

Answer:

THE VELOCITY TO STRIKE THE GROUND IS

-332.38 m/s

I HOPE IT HELPS ❤️❤️❤️

21. Choose the letter 1. Which of the following statement is NOT true? a. the horizontal acceleration is always 0 b. vertical acceleration of a projectile is constants c. horizontal and vertical components of projectile are dependent on each other d. Horizontal and vert. horizontal and vertical components of projectile are independent on each other 5.Which of the following statement is true about the vertical motion of projectiles a. the vertical component of a projectiles velocity is constant b. the vertical component of a projectiles velocity is changing c. the vertical component of a projectiles velocity is constant value of 9.8m/s d. the vertical component of a projectiles velocity is changing at a changing rate 6.Which of the following is independent of the horizontal component of a projectile velocity a. its range b. the vertical component c. time d. none of these 7.What component is the velocity of a ball at the very peak of the ball path if it is thrown into the air at some angles a. entirely vertical b. entirely horizontal c. not enough information d. both vertical and horizontal 8.What do you call the path of the a sepak takraw that is kicked from the height of 2m a. circular b. hyperbolic c. linear d. parabolic 9.What is the vertical acceleration of the ball after 1 second after a baseball is thrown horizontally a.0 b.1m/s2 c.9.8m/s2 d. -9.8m/s2 10.How fast is a stone is fallen from a clip at 6 seconds? a.9.8m/s b.19.6m/s c. 58.8m/s d.0m/s

[tex]answer[/tex]

1.A

2.B

3.C

4.A

5.B

6.D

7.A

8.B

9.A

10.B

[tex]hope \: it \: helps[/tex]

22. ACTIVITY Direction: Given the angle and initial velocity, calculate the horizontal and vertical components of projectile. Angle (2) Vertical Component Initial Velocity Horizontal Component 15° 2.59 m/s 10 m/s 9.66 m/s 30° 10 m/s 45° 10 m/s 60° 10 m/s 75° 10 m/s​

Answer:

Yey✨ sikat Nako nandito si Tyrone Tiaga✨✨

Pa Wash awt idol

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23. Which of the following statements is TRUE of the horizontal motion of projectiles?A. The range traveled by a projectile is dependent upon the time of flight and vertical velocityB. The horizontal velocity of a projectile affects the maximum height reached by a projectile. C. The horizontal and vertical velocities of a projectile are completely independent of each otherD. The horizontal velocity of a projectile changes by 9.8m/s Previous​

Answer:

C:The horizontal and vertical velocities of a projectile are completely independent of each other

Explanation:

The horizontal and vertical motions of a projectile are independent, meaning they do not affect each other.

not sure about my answer but i wish this helps

Her the same time as part of the poet crossword puzzles are asked to come God bless you

24. A projectile thrown upward at an angle θ with the horizontal covers a range of 12 m in a span of 5.0 s. Find the (a) horizontal and vertical components of the initial velocity, (b) angle of projection θ, (c) initial velocity, and (d) maximum height attained by the projectile

Answer:

a. The horizontal and vertical components of the initial velocity are 2.4m/s and 24.5 m/s respectively

b. The angle of projection is 84.402°

c. The initial velocity is 24.62 m/s

d. The maximum height attained by the projectile is 30.63 meters

Explanation:

In this problem, we will be using the concept of projectile motion to solve the following requirements.

Formula to use

a. Time of flight

t = (2V₀ sinθ) /g              equation 1

b. Range

R = (V₀² sin2θ)/ g           equation 2

c. Maximum height

H = (V₀² sin²θ)/ 2g          equation 3

where

V₀     is the initial velocity, the unit is in m/s

θ       is the projection angle

g       is the acceleration due to gravity, 9.8 m/s²

Given information

R = 12m     (range)

t = 5s         (time of flight)

a. Vx = ?, Vy = ?

b. θ = ?

c. V₀ = ?

d. H = ?

Solving the problem

a. Let us use equation 1 and 2 to solve the initial velocity and projection angle

t = (2V₀ sinθ) /g              eq.1

R = (V₀² sin2θ)/ g           eq.2

Use eq.1 then substitute it to eq.2:

5 = (2V₀ sinθ) /9.8              

Simplify the equation, we have:

V₀ sinθ = 24.5

V₀ = 24.5/sinθ      equation 1'

Equation 2:

R = (V₀² sin2θ)/ g

12 = (V₀² sin2θ)/ 9.8

V₀² sin2θ = 117.6

(24.5/sinθ)² · sin2θ = 117.6

sin2θ / (sinθ)² = 0.196

Take note that sin2θ = 2sinθ cosθ

2sinθ cosθ/(sinθ)² = 0.196

2cosθ/sinθ = 0.196

2cot θ = 0.196

Solve for θ :

θ = cot⁻¹ (0.196/2)

θ = 84.402°

Now, let us solve the initial velocity, V₀ using equation 1':

V₀ = 24.5/sinθ

V₀ = 24.5/sin 84.402

V₀ = 24.62 m/s

Now, we can solve for the horizontal and vertical components:

Horizontal component:

Vx = V₀ cosθ

Vx = 24.62 cos 84.402

Vx = 2.4 m/s

Vertical component:

Vy = V₀ sin θ

Vy = 24.62 sin 84.402

Vy = 24.5 m/s

Finally, let us solve the maximum height using the formula

H = (V₀² sin²θ)/ 2g          

then, substitute the given data

H = 24.62² (sin 84.402)²/ 2(9.8)

H = 30.63 meters

To learn more, just click the following links:

Description of projectile motion

       https://brainly.ph/question/2662013

Examples of projectile motion

       https://brainly.ph/question/107266

#LetsStudy

25. Which of the following statements is true about the vertical motion of projectiles?a. The vertical component of a projectile's velocity is constant.b. The vertical component of a projectile's velocity is changing.c. The vertical component of a projectile's velocity is a constant value of 9.8 m/s.d. The vertical component of a projectile's velocity is changing at a changing rate. ​

Answer:

The horizontal velocity of a projectile is constant (a never changing in value), There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down, The vertical velocity of a projectile changes by 9.8 m/s each second, The horizontal motion of a projectile is independent of its vertical motion.

Explanation:

so the c didn't say that the 9.8 m/s didnt change every seconds/second so its letter D.

Answer:

Question:Which of the following statements is true about the vertical motion of projectiles?

Answer:

C.The vertical component of a projectiles velocity is a constant value of 9.8 m/s

Hope it helps

26. horizontal velocity component of a projectile formula​

Answer:

As a result, we have only one component of initial velocity - Vx = V, whereas Vy = 0. Horizontal distance can be expressed as x = V * t . Vertical distance from the ground is described by the formula y = – g * t² / 2 , where g is the gravity acceleration and h is an elevation.

Explanation:

27. what is the horizontal component of the velocity of a projectile if it is through air with zero horizontal acceleration?​

Answer:

Accelaration is a vector quantity. That is, it has a magnitude as well as a direction.

We have something called gravity. It's simply a force due to the mass of a body. That is, if two bodies has some mass then a force will exist between them. That force will attract each other. So, if you have two balls A and B having some mass, then A will pull B towards itself and likewise B will pull A towards itself.

In our case, the entire earth plays the role of ball A, and the other body ( say a ball with which we're playing) plays the role of ball B. So the earth will pull our ball towards itself and likewise the ball will also. But because the mass of the earth is much more than that of the ball so ball moves towards it.

Now, whenever we represent the earth on paper in Physics, we draw it as a flat surface. So pulling towards it would mean pulling vertically downwards.

We take the earth as a reference and make x axis along the surface of the earth and y axis, perpendicular to it.

So, we have only one acceleration and which is towards the earth. And 'towards earth' means vertically downwards. So there is no question of horizontal acceleration. The earth is not pulling the ball horizontally.

Hence in projectiles, there is no horizontal acceleration i.e horizontal acceleration is zero.

There is horizontal acceleration (against the direction of motion) in projectile motion. A projectile (like a gun shell) will decelerate due to atmospheric drag. The phenomenon has been measured since a long time (using the ballistic pendulum: ).

The deceleration is related to the ballistic coefficient . A projectile with a high ballistic coefficient (heavy-low drag), like a heavy shell, will have little deceleration, while a projectile with low ballistic coefficient (light-high drag), like a air balloon that you would throw, will i

It turns out you can analyze projectile motion problems by looking at what happens in the vertical direction (acceleration due to gravity) and horizontal direction (no acceleration whatsoever) and merging the results. This is called superposition.

Remember Newton’s first law of motion? You probably learned it on the very first day of Physics class.

Paraphrased, “A body in motion, tends to remain in motion, unless acted upon by an external force.”

Once the projectile is released, there is no horizontal force (neglecting air resistance). No Force => No Acceleration. This follows from Newton’s second law, expressed in an equation as:

F = mA

If F(horizontal) is zero, A(horizontal) is zero. It is as simple and wonderful as that!

28. in projectile the horizontal component of the velocity is​

Answer:

In a projectile motion, if we neglect the air friction, the only force that acts is the gravitational force which is in the downward direction. Thus, the horizontal component of velocity remains constant as their is no force in the horizontal direction.

29. at what point in its path is the horizontal component of the velocity of a projectile the smallest

Answer:

At the top.

Explanation:

A projectile path has a vertical and horizontal component to it or 2d kinematics. But gravity is acting on the projectile the whole time so it is going to stop it from going up and bring it back to the ground at its max ordinance. At that instant the velocity would be 0.

ctto.

30. describe the horizontal and vertical components of a velocity of a projectile

The horizontal velocity of a projectile is constant (a never changing in value), There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down, The vertical velocity of a projectile changes by 9.8 m/s each second, The horizontal motion of a projectile is independent of its vertical motion