an anti-aircraft shell is fired vertically upward with a muzzle velocity of 488 m\s.Determine

1. an anti-aircraft shell is fired vertically upward with a muzzle velocity of 488 m\s.Determine

 v^2 = 2gs 
s = v^2 / 2g and g = 9.8 m/s^2 

s = 488 ^2/(2 * 9.8) 
= 12150 m approximately 
i.e a bit over 12km

2. An antiaircraft shell is fired vertically upward with an initial speed of 500m/s. Neglecting friction, compute the maximum height it can reach. The time taken to reach that height, and the instantaneous velocity at the end of 60s. When will its height be 10km?

Answer:

120s UTOT ANG ANSWER HAHAAHAHA

Explanation:

Because of the gravity of the earth

3. . When you fire a gun at anangle, what happens to its upward vertical velocity?​

Answer:

As the projectile moves upwards it goes against gravity, and therefore the velocity begins to decelerate. Eventually the vertical velocity will reach zero, and the projectile is accelerated downward under gravity immediately. Once the projectile reaches its maximum height, it begins to accelerate downward.

4. the acceleration of the ball thrown vertically upward is a uniform or not​

Answer:

The direction of the ball's motion remains upward on its way up, but its speed decreases down to zero at the very top of its motion. Throughout the ball's travel upward and then downward, the acceleration is constant, 9.8m/s2 (the acceleration due to gravity)

Answer:

the acceleration of the ball thrown vertically upward is a uniform or not

5. A cannon fires a shell at a fixed angle above the horizontal.Which one of the following quantities is the same throughout the shell's flight?(ignore the effects of air friction)A. speedB. accelerationC. velocityD. vertical velocity​

Answer:

D.vertical velocity

Explanation:

#Carry on learning

6. Which of the following statements is true for a ball thrown vertically upward?

Answer:

the ball has a negative acceleration on the way up and a positive acceleration on the way down.

Explanation:

hope its help :)

7. a volleyball is tossed vertically upward

Answe  r:

then it moves down

Explanation:

Answer:

Tossing an object upward or a free fall of an object is a special type of projectile motion. In tossing a volleyball vertically upward gives an initial velocity and as it moves upward, the velocity decreases until the vertical velocity is zero.

8. a volleyball is tossed vertically upward brainly

Answer:

TRUE

Explanation:

CORRECT ME IF I AM WRONGYes
Jdndjsnskdnksndndjsnd

9. An arrow is shot vertically upward at a speed as high as 100m/s

Answer:

[tex] \color{cyan}\rule{1pt}{999999pt}[/tex]

[tex] \color{cyan}\rule{1pt}{999999pt}[/tex]

Explanation:

[tex] \color{cyan}\rule{1pt}{999999pt}[/tex]

[tex] \color{cyan}\rule{1pt}{999999pt}[/tex]

10. What happens to the velocity of an object if it is thrown vertically upward?

Answer:

When you throw an object upwards, it will eventually fall back to the ground under the earth's gravity. ... An object that is thrown vertically upwards decelerates under the earth's gravity. Its speed decreases until it attains a maximum height, where the velocity is zero.

Explanation:

Hope its help

Pa brainleast tnx

Stay safe always

11. A shell leaves a mortar with a muzzle velocity of 500 ft/s directed upward at 60 degrees with the horizontal. Determine the position of the shell and its resultant velocity 20 s after firing. how high will it rise?

Answer:

a. The position of the shell is 5470.79 ft.

b. The resultant velocity is 327.13 ft/s

c. The maximum height is 2911.49 ft.

Explanation:

In this case, the problem is about Projectile. It is any body with an initial velocity that follows a path or trajectory which is affected by gravitational acceleration and air resistance.

A. Formula to be used:

1. To find the position of the shell, we will use the formula

[tex]r=\sqrt{x^2+y^2}[/tex]

where

[tex]r[/tex]          is a vector for the projectile position, it is the resultant

          distance of the projectile or position (magnitude), unit is

          in meters (m) or feet (ft.)            

[tex]x[/tex]          is the horizontal position of the projectile from the

           starting point to the landing position, unit is in meters

           (m) or feet (ft.)

[tex]y[/tex]          is the vertical position of the projectile, unit is also in

          meters (m) or feet (ft.)            

2. To find resultant velocity of the shell, we will use the formula

[tex]v=\sqrt{v_{x}^2+v_{y}^2 }[/tex]

where

[tex]v[/tex]          is the resultant velocity, unit is in m/s or ft/s

[tex]v_{x}[/tex]        is the horizontal velocity affecting the projectile's speed,

          unit is also in m/s or ft/s

[tex]v_{y}[/tex]       is the vertical velocity also affecting the projectile's

          speed of the object, unit is also in m/s or ft/s

3. To find how high will the shell will rise (the maximum height of the shell), we will use the formula

[tex]y_{max} =\frac{V_{0}^2Sin^2\theta }{2g}[/tex]

where

[tex]y_{max}[/tex]      is the maximum height the shell will rise, unit is in

            meter or feet

[tex]V_{0}[/tex]          is the initial velocity, unit is in m/s or ft/s

θ           is the angle created with the positive x-axis, unit is in

             degrees

[tex]g[/tex]           is the acceleration due to gravity, we will use 32.2 ft/[tex]s^2[/tex]

            here

B. For the given information

[tex]V_{0}[/tex] = 500 ft/s

θ = 60°

t = 20 s

C. Solving the problem

1. First, we will solve the position of the shell. To solve this, we need to add another formula for [tex]x[/tex] and [tex]y[/tex]:

a. [tex]x=V_{0} cos\theta(t)[/tex]

   [tex]y=(V_{0} Sin\theta)t-\frac{1}{2} gt^2[/tex]

b. Let's solve first [tex]x[/tex] and [tex]y[/tex]:

[tex]x=V_{0} cos\theta(t)[/tex]                                          

[tex]x=(500 cos60)(20s)[/tex]  

x = 5000 ft.                        

[tex]y=(V_{0} Sin\theta)t-\frac{1}{2} gt^2[/tex]

[tex]y=(500ft/s)( Sin60)(20s)-\frac{1}{2} (32.2ft/s^2)(20)^2[/tex]

y = 2220.25 ft.

c. Now, we are ready to solve for the resultant position, [tex]r[/tex]

[tex]r=\sqrt{x^2+y^2}[/tex]

[tex]r=\sqrt{5000^2+2220.25^2}[/tex]

r = 5470.79 ft.

2. For the second question, we will solve for the resultant velocity, [tex]v[/tex]. But before we solve this, another formula should be added: [tex]v_{x}[/tex] and [tex]v_{y}[/tex]

a. [tex]v_{x} =V_{0} cos\theta[/tex]

   [tex]v_{y} =V_{0} sin\theta-gt[/tex]

b. Let us solve first for [tex]v_{x}[/tex] and [tex]v_{y}[/tex]

[tex]v_{x} =V_{0} cos\theta[/tex]                   [tex]v_{y} =V_{0} sin\theta-gt[/tex]

[tex]v_{x} =(500ft/s) cos60[/tex]      [tex]v_{y} =(500ft/s) (sin60)-(32.2ft/s^2)(20s)[/tex]

[tex]v_{x}[/tex] = 250 ft/s                  [tex]v_{y}[/tex] = - 210.987 ft/s

c. Now, we are ready to solve for the resultant velocity. Use the equation for resultant velocity and substitute the values of [tex]v_{x}[/tex] and [tex]v_{y}[/tex]

[tex]v=\sqrt{v_{x}^2+v_{y}^2 }[/tex]

[tex]v=\sqrt{(250ft/s)^2+(-210.987ft/s)^2}[/tex]

v = 327.13 ft/s

3. For the last question, we are to solve for the maximum height the shell will rise.

Using the formula for maximum height, [tex]y_{max}[/tex] then substitute the given information

[tex]y_{max} =\frac{V_{0}^2Sin^2\theta }{2g}[/tex]

[tex]y_{max} =\frac{(500ft/s)^2(Sin60)^2 }{2(32.2ft/s^2)}[/tex]

[tex]y_{max}[/tex] = 2911.49 ft.

Therefore, the position of the shell is 5470.79 ft., the resultant velocity is 327.13 ft/s and the maximum height is 2911.49 ft.

To learn more about projectile, just click the following links:

Definition of a projectile motion

       https://brainly.ph/question/100593

Examples of projectile motion

       https://brainly.ph/question/107266

Two components of projectile motion

       https://brainly.ph/question/99671

#LetsStudy

12. When a stone is thrown vertically upwards?

PA BRAINLIEST PO

Answer:

When the stone is at a height equal to half of its maximum height its speed will be 10 m/s, then the maximum height attained by the stone is (Take g =10 m/s2)

13. A projectile is fired vertically upward from the ground with an initial velocity of 120m/s . if s meters is the height of the projectile above the ground t seconds after being fired, express s in terms of t, under the assumption that the only force acting on the projectile is attributed to the acceleration due to gravity.

Answer:

yan po

Explanation:

paki brainliest napo plss.

14. a sepak takraw ball is hit vertically upward

Answer:

it had great force

Explanation:

15. An artillery shell has a mass of 20kg. The shell is fired from a gun with a force of 48N. What is the acceleration of the shell​

Answer:

a=f/m

  =48N/20kg

  =2.4 m/s2

16. a sepak takraw ball is hit vertically upward by a player

Answer:

-9.8 per square second

Explanation:

Pa-Brainliest po!Hope it helps

17. Why does the production of the Canadian anti-aircraft tank "Skink" cancelled? in WW2?

Answer:

The Second World War was a defining event in Canadian history, transforming a quiet country on the fringes of global affairs into a critical player in the 20th century's most important struggle. Canada carried out a vital role in the Battle of the Atlantic and the air war over Germany, and contributed forces to the campaigns of western Europe beyond what might be expected of a small nation of then only 11 million people.

Explanation:

18. Which is the softest part of the fired shell?

Answer:The part of a firearm that loads fires, and ejects a cartridge. Includes

lever action, pump action, bolt action, and semi-automatic. The first three are

found in weapons that fire a single shot. Firearms that can shoot multiple

rounds ("repeaters") include all these types of actions, but only the semiautomatic does not require manual operation between rounds. A truly

"automatic" action is found on a machine gun

Explanation:

19. a volleyball is tossed vertically upward with an initial velocity

Answer:

it falls and moves vertically fast

Explanation:

Answer:

TRUE

Explanation:

HOPE IT HELPS

CORRECT ME IF I AM WRONG

20. 'How long will it take a shell fired from a cliff at an initial velocity of 800m/s at an angle 30 deg below the horizontal to reach the ground 150m Read more: On level ground a shell is fired with an initial velocity of 80 m/s at 60 degrees above the horizontal and feels no appreciable air resistance? - Find the horizontal and vertical components of the shell`s velocity.

 ito yung para sa second question mo
Horizontal component = (cos 30 degrees) x 80 = 69.28 m/s
Vertical component = (sin 30 degrees) x 80 = 40 m/s
 ito yung sa una.. sorry akala ko kasi nagkamali k type
two pla questions mo


H=H0+ V0(t)+ 1/2 G (t^2)

H0= 150m
V0= -800 sin 30 degrees = -400m/s
G= 9.8 m/s^2
t= time
substitute
H=150-400t- 4.9t^2
use quadratic eqtn
t= -400 +/- squareroot of (400^2+ 4*4.9*150)/-9.8
t= -400+/- squareroot of (160000+2940)/-9.8
t= 82.01 sec
t= 0.3733 sec
answer it will take 0.3733 sec for the shell to reach the ground

21. ano ang a ball is thrown vertically upward

answer isss force thats too easy

22. A ball thrown vertically upward at 80 m/s will reach the ground in

the ball will reach 50m/s in the ground

23. A football is kicked vertically upward from the ground and a student

projectile motion and it moves away from the person who kick the ball. the more you force you give the far it will go. 95 degree is the proper angle to be use.

24. how would you know where a. the upward vertical velocity the greatest b. the horizontal velocity the greatest c. where the vertical velocity is zero d. the downward vertical velocity the least e. the upward vertical velocity the least

A the upward vertical velocity the greatest

Answer:

A. the upward vertical velocity the greatest

25. A ball is thrown vertically upward with a speed of 35.0 m/s

Answer:

hello its me you looking for

Explanation:

happy new year

26. A mortar at an angle of 30° with the ground fires a shell with a muzzle velocity of 196 meters per second.The maximum height of the shell is approximately​

Answer:

70 degrees sorry if it is wrong

27. A body is thrown vertically upward maximum height is reached and then it will have

Answer:

it will have zero velocity and finite acceleration

Explanation:

natapon ko sarili ko

28. A rocket is fired vertically upward with an initial velocity of 20 m/s. What is the maximum height of the rocket?

2gdy = Vf^2 - Vo^2
2(9.8m/s^2)dy = 20m/s^2 - 0m/s^2
19.6m/s^2(dy) = 400m^2/s^2
dy = 400m^2/s^2 / 19.6m/s^2

dy=20.408 m

29. Released from rest. thrown vertically downward. thrown vertically upward. which of the balls will reach the ground with greatest speed?

the one you threw downward :)

That would be thrown vertically downward. This is because an additional force is exerted and this force accelerates the object downward.

30. A shell fired 65m from a cliff and reaches the water with a final vertical velocity of 800m/s. How much time had elapsed before he reached the water?

0.08125s

Explanation:

T=D÷V

T=65÷800m/s

T=0.08125s